Calculating K_{c} from a known set of equilibrium concentrations seems pretty clear. You just plug into the equilibrium expression and solve for K_{c}.

Calculating equilibrium concentrations from a set of initial concentrations takes more calculation steps. In this type of problem, the K_{c} value will be given

The best way to explain is by example. Just in case you are not sure, the subscripted zero, as in [H_{2}]_{o}, means the initial concentration.

**Example #1:** Given this equation:

H_{2}+ I_{2}⇌ 2HI

Calculate all three equilibrium concentrations when [H_{2}]_{o} = [I_{2}]_{o} = 0.200 M and K_{c} = 64.0.

**Solution:**

1) The solution technique involves the use of what is most often called an ICEbox. Here is an empty one:

[H _{2}][I _{2}][HI] Initial Change Equilibrium The ChemTeam hopes you notice that I, C, E are the first initials of Initial, Change, and Equilibrium.

2) Now, let's fill in the initial row. This should be pretty easy:

[H _{2}][I _{2}][HI] Initial 0.200 0.200 0 Change Equilibrium The first two values were specified in the problem and the last value ([HI] = 0) come from the fact that the reaction has not yet started, so no HI could have been produced yet.

3) Now for the change row. This is the one that causes the most difficulty in understanding:

[H _{2}][I _{2}][HI] Initial 0.200 0.200 0 Change −x −x +2x Equilibrium The minus sign comes from the fact that the H

_{2}and I_{2}amounts are going to go down as the reaction proceeds.x signifies that we know some H

_{2}and I_{2}get used up, but we don't know how much. What we do know is that an EQUAL amount of each will be used up. We know this from the coefficients of the equation. For every one H_{2}used up, one I_{2}is used up also.The positive signifies that more HI is being made as the reaction proceeds on its way to equilibrium.

The two is important. HI is being made twice as fast as either H

_{2}or I_{2}are being used up.In fact, always use the coefficients of the balanced equation as coefficients on the "x" terms.

In problems such as this one, never use more than one unknown. Since we have only one equation (the equilibrium expression) we cannot have two unknowns.

4) The equilibrium row should be easy. It is simply the initial conditions with the change applied to it:

[H _{2}][I _{2}][HI] Initial 0.200 0.200 0 Change −x −x +2x Equilibrium 0.200 − x 0.200 − x 2x

5) We are now ready to put values into the equilibrium expression. For convenience, here is the equation again:

H_{2}+ I_{2}⇌ 2 HIand the equilibrium expression is:

[HI] ^{2}K _{c}=––––––– [H _{2}] [I_{2}]

6) Plugging values into the expression gives:

(2x) ^{2}64.0 = ––––––––––––––––– (0.200 − x) (0.200 − x)

Two points need to be made before going on:

1) Where did the 64.0 value come from? It was given in the problem.

2) Make sure to write (2x)^{2}and not 2x^{2}. As you well know, they are different. This mistake happens a LOT!!

6) Both sides are perfect squares (done so on purpose), so we square root both sides to get:

2x 8.00 = ––––––––– 0.200 − x From there, the solution should be easy and results in x = 0.160 M.

7) This is not the end of the solution since the question asked for the equilibrium concentrations, so:

[H_{2}] = 0.200 − 0.160 = 0.040 M

[I_{2}] = 0.200 − 0.160 = 0.040 M

[HI] = (2) (0.160) = 0.320 M

8) You can check for correctness by plugging back into the equilibrium expression:

(0.320) ^{2}K _{c}=–––––––––––– (0.040) (0.040) K

_{c}= 64.0Yay!

In the second example, the quadratic formula will be used.

**Example #2:** Given this equation:

PCl_{3}+ Cl_{2}⇌ PCl_{5}

Calculate all three equilibrium concentrations when K_{c} = 16.0 and [PCl_{5}]_{o} = 1.00 M.

**Solution:**

1) Here is the completed ICEbox:

[PCl _{3}][Cl _{2}][PCl _{5}]Initial 0 0 1.00 Change +x +x −x Equilibrium x x 1.00 − x

2) The equilibrium expression is:

[PCl _{5}]K _{c}=––––––––––– [PCl _{3}] [Cl_{2}]Substituting gives:

1.00 − x 16.0 = ––––––– (x) (x)

3) After suitable manipulation (which you can perform yourself), we arrive at this quadratic equation in standard form:

16x^{2}+ x − 1 = 0

4) Using the quadratic formula:

$x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$and

a = 16, b = 1 and c = −1

we obtain:

x = 0.2207

To three sig figs, 0.221

5) Please notice that the negative root was dropped, because −b turned out to be −1. The answer obtained in this type of problem CANNOT be negative. Why?

Because we are dealing with the amount of a physical substance in mol / L. Amounts of substances are always represented with positive numbers. An amount of a substance with physical reality cannot be represented with negative numbers.

5) Determination of the equilibrium amounts and checking for correctness by inserting back into the equilibrium expression is left to the student. The answer you get will not be exactly 16, due to errors introduced by rounding.

The third example will be one in which both roots give positive answers. The question then becomes how to determine which root is the correct one to use.

**Example #3:** Given this equation:

COCl_{2}⇌ CO + Cl_{2}

Calculate all three equilibrium concentrations when K_{c} = 0.680 with [CO]_{o} = 0.500 and [Cl_{2}]_{o} = 1.00 M.

**Solution:**

1) Here is the completed ICEbox:

[COCl _{2}][CO] [Cl _{2}]Initial 0 0.500 1.00 Change +x −x −x Equilibrium x 0.500 − x 1.00 − x

2) The equilibrium expression is:

[CO] [Cl _{2}]K _{c}=––––––––– [COCl _{2}]Substituting gives:

(0.5 − x) (1 − x) 0.680 = ––––––––––––– x

3) After some manipulation (left to the student), we arrive at this quadratic equation, in standard form:

x^{2}− 2.18x + 0.5 = 0

4) Using a quadratic equation solver, we wind up with this:

x = (2.18 ± 1.66) / 2

5) Both roots yield positive values, so how do we pick the correct one?

The answer lies in the fact that x is not the final answer, whereas (0.5 − x) is. It is the term (0.5 − x) which must be positive.So the root of 1.92 is rejected in favor of the 0.26 value and the three equilibrium concentrations can be calculated.

The day is saved!!

**Example #4:** Given this equation:

H_{2}+ Br_{2}⇌ 2HBr

Calculate all three equilibrium concentrations when 0.500 mole each of H_{2} and Br_{2} are mixed in a 2.00 L container and K_{c} = 36.0.

**Comment:**

This problem has a slight trick in it. Notice that moles are given and volume of the container is given. However, the calculations must be done in molarity. So you must divide 0.500 by 2.0 to get 0.250 mol/L. That is the number to be used.

In my classroom, I used to point this out over and over, yet some people seem to never hear. I promise them I will test it and when I do, many people use 0.500 for their calculation, not 0.250.

I hope you don't get caught in the same mistake.

**Solution:**

1) We will use an ICEbox. Here is the initial row, filled in:

[H _{2}][Br _{2}][HBr] Initial 0.250 0.250 0 Change Equilibrium Remember, the last value of zero come from the fact that the reaction has not yet started, so no HBr could have been produced yet.

2) Now for the change row:

[H _{2}][Br _{2}][HBr] Initial 0.250 0.250 0 Change −x −x +2x Equilibrium The minus sign tends to mess people up, even after it is explained over and over. It is associated with the substances being used up as the reaction goes to equilibrium. Some people never seem to figure that something (in this case, H

_{2}and Br_{2}) are going away and some new stuff (the HBr) is comming in.x signifies that we know some H

_{2}and Br_{2}get used up, but we don't know how much. What we do know is that an EQUAL amount of each will be used up. We know this from the coefficients of the equation. For every one H_{2}used up, one Br_{2}is used up also.This also messes up a lot of people. I think it is because they do not have a good idea in their brain about what is happening during the chemical reaction. They have a hard time with the concept that the H

_{2}splits into two separate H and the Br_{2}splits into two Br. The each of the two H and two Br hook together to make two different HBr molecules.Now, I can just see some of you sitting there saying, "Geez, what a wasted paragraph." No way man, there are people who DO NOT GET IT. Those people are in your class and you know who they are. Go give them a bit of help.

The two is important. HI is being made twice as fast as either H

_{2}or I_{2}are being used up.In fact, always use the coefficients of the balanced equation as coefficients on the "x" terms. Even if you don't understand why, memorize the idea that the coefficients attach on front of each x. Another way: the coefficient of each substance in the chemical equation becomes the coefficient of its 'x' in the change row of the ICEbox.

3) The equilibrium row is:

[H _{2}][Br _{2}][HBr] Initial 0.250 0.250 0 Change −x −x +2x Equilibrium 0.250 − x 0.250 − x 2x

4) Now we are are ready to put values into the equilibrium expression. For convenience, here is the equation again:

H_{2}+ Br_{2}⇌ 2 HBr

5) The equilibrium expression is:

[HBr] ^{2}K _{c}=–––––––– [H _{2}] [Br_{2}]

6) Plugging values into the expression gives:

(2x) ^{2}36.0 = –––––––––––––––––– (0.250 − x) (0.250 − x)

7) Two points need to be made before going on:

1) Where did the 36.0 value come from? It was given in the problem.

2) Make sure to write (2x)^{2}and not 2x^{2}. As you well know, they are different. This mistake happens a LOT!! This mistake happens on the test even after 10-15 mentions of it in class.

8) Both sides are perfect squares (done so on purpose), so we square root both sides to get:

2x 6.00 = ––––––––– 0.250 − x

9) From there, the solution should be easy. You can check for correctness by plugging back into the equilibrium expression.

This example will involve the use of the quadratic formula.

**Example #5:** Given this equation:

H_{2}+ Cl_{2}⇌ 2HCl

Calculate all three equilibrium concentrations when K_{c} = 20.0 and [H_{2}]_{o} = 1.00 M and [Cl_{2}]_{o} = 2.00 M.

**Solution:**

1) Here is the completed ICEbox:

[H _{2}][Cl _{2}][HCl] Initial 1.00 2.00 0 Change −x −x +2x Equilibrium 1.00 − x 2.00 − x 2x

2) The equilibrium expression is:

[HCl] ^{2}K _{c}=–––––––– [H _{2}] [Cl_{2}]

3) Substituting gives:

(2x) ^{2}20.0 = ––––––––––––––––– (1.00 − x) (2.00 − x)

4) After suitable manipulation (which you can perform yourself), we arrive at this quadratic equation in standard form:

16x^{2}− 60x + 40 = 0

5) Using the quadratic formula, we obtain:

x = 0.867.

6) In this problem, note that −b equals −(−60). This means both roots will probably be positive. Which one should you check first?

Always check the minus root first. it will give the smaller answer and that's usually what you want. In this problem, the larger root gives answer of 2.88, which leads to impossible results.

7) Determine the equilibrium concentrations and then check for correctness by inserting back into the equilibrium expression.

**Example #6:** 0.850 mol each of N_{2} and O_{2} are introduced into a 15.0 L flask and allowed to react at constant temperature. NO is the sole product. What are the concentrations of all three chemical species after the reaction has come to equilibrium? The K_{c} was determined in another experiment to be 0.0125.

**Solution:**

1) Write the chemical equation:

N_{2}+ O_{2}⇌ 2NO

2) Determine initial concentrations:

[N_{2}]_{o}= [O_{2}]_{o}= 0.850 mol / 15.0 L = 0.0567 M

3) Set up an ICEbox:

[N _{2}][O _{2}][NO] Initial 0.0567 0.0567 0 Change −x −x +2x Equilibrium 0.0567 − x 0.0567 − x 2x

4) Write the equilibrium constant expression, substitute values and solve:

K_{c}= [NO]^{2}/ ([N_{2}] [O_{2}])0.0125 = (2x)

^{2}/ [(0.0567 - x) (0.0567 - x)]0.112 = 2x / (0.0567 - x)

2x = 0.00635 - 0.112x

2.112x = 0.00635

x = 0.00301 M

5) Determine the equilibrium concentrations:

[N_{2}] = 0.0567 − 0.00301 = 0.0534 M

[O_{2}] = 0.0567 − 0.00301 = 0.0534 M

[NO] = (2) (0.00301) = 0.00602 M

6) These values can be checked by inserting them back into the K_{c} equation:

0.0125 ?=? (0.00602)^{2}/ (0.0534)^{2}To a reasonable amount of error (caused by rounding), the values are shown to be correct.

**Example #7:** Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, the reaction below can proceed to a measurable extent.

NAt 3000 K, the reaction above has K_{2}(g) + O_{2}(g) ⇌ 2NO(g)

**Solution:**
1) The ICEbox with just the initial conditions:

[NO]

_{o}---> 0.3000 mol / 2.000 L = 0.1500 M

[N _{2}][O _{2}][NO] Initial 0 0 0.1500 Change Equilibrium

2) Change:

[N _{2}][O _{2}][NO] Initial 0 0 0.1500 Change +x +x −2x Equilibrium Remember, the change is based on the stoichiometry of the reaction. For every two NO that decompose, one N

_{2}and one O_{2}are formed.

3) Equilibrium:

[N _{2}][O _{2}][NO] Initial 0 0 0.1500 Change +x +x −2x Equilibrium x x 0.1500 − 2x

4) Write the equilibrium expression, put values in, and solve:

[NO] ^{2}K _{c}=–––––––– [N _{2}] [O_{2}]

(0.1500 − 2x) ^{2}0.0153 = ––––––––––– (x) (x)

(0.1500 − 2x) 0.1237 = ––––––––––– x x = 0.0706 M

5) [NO] at equilibrium:

0.1500 − [(2) (0.0706)] = 0.0096 M

**Example #8:** At 2200 °C, K_{p} = 0.050 for the reaction;

N_{2}(g) + O_{2}(g) ⇌ 2NO(g)

What is the partial pressure of NO in equilibrium with N_{2} and O_{2} that were placed in a flask at initial pressures of 0.80 and 0.20 atm, respectively?

**Solution:**

Comment: the calculation techniques for treating K_{p} problems are the exact same techniques used for K_{c} problems. For the same reaction, the K_{p} and K_{c} values can be different, but that play no role in how the problem is solved.

1) An ICEbox with the initial pressures:

P _{N2}P _{O2}P _{NO}Initial 0.80 0.20 0 Change Equilibrium Notice that pressures are used, not concentrations.

2) Change:

P _{N2}P _{O2}P _{NO}Initial 0.80 0.20 0 Change −x −x +2x Equilibrium

3) Equilibrium conditions:

P _{N2}P _{O2}P _{NO}Initial 0.80 0.20 0 Change −x −x +2x Equilibrium 0.80 − x 0.20 − x 2x

4) Write the equilibrium constant expression, substitute values into it, and solve:

(P _{NO})^{2}K _{p}=–––––––– (P _{N2}) (P_{O2})

(2x) ^{2}0.050 = –––––––––––––––– (0.80 − x) (0.20 − x)

4x ^{2}0.050 = ––––––––––– 0.16 − x + x ^{2}0.008 − 0.05x + 0.05x

^{2}= 4x^{2}3.95x

^{2}+ 0.05x − 0.008 = 0

5) A quadratic equation solver is used. The negative root is discarded. The answer is determined to be:

x = 0.039 atmTherefore, P

_{NO}= 0.078 atm

**Example #9:** Consider the reaction:

2HI(g) ⇌ H_{2}(g) + I_{2}(g)

at 620 °C where K = 1.63 x 10¯^{3}. Given that [H_{2}]_{o} = 0.300 M, [I_{2}]_{o} = 0.150 M and [HI]_{o} = 0.400 M, calculate the equilibrium concentrations of HI, H_{2}, and I_{2}.

**Solution:**

1) ICEbox with initial conditions:

[HI] [H _{2}][I _{2}]Initial 0.400 0.300 0.150 Change Equilibrium

2) The question becomes "Which way will the reaction go to get to equilibrium? Will it go to the right (more H_{2} and I_{2})? Or, will it go to the left (more HI)? To answer that, we use a concept called the reaction quotient:

[H _{2}]_{o}[I_{2}]_{o}Q = –––––––– [HI] _{o}^{2}The reaction quotient is based on the initial values only, before any reaction takes place.

3) Calculate the value of Q:

(0.300) (0.150) Q = –––––––––––– (0.400) ^{2}Q = 0.28125

4) Now, we compare Q to K_{c}: Is Q greater than, lesser than, or equal to K_{c}?

0.28125 is greater than 0.00163. This means that the equilibrium will shift to the left, with the goal of obtaining 0.00163 (the K_{c}). The amounts of H_{2}and I_{2}will go down and the amount of HI will go up. The value of Q will go down until the value for K_{c}is arrived at.

5) We can now write the rest of the ICEbox . . .

[HI] [H _{2}][I _{2}]Initial 0.400 0.300 0.150 Change +2x −x −x Equilibrium 0.400 + 2x 0.300 − x 0.150 − x

6) . . . and insert values in the equilibrium expression:

(0.300 − x) (0.150 − x) 0.00163 = –––––––––––––––––– (0.400 + 2x) ^{2}

7) Algebra!

x ^{2}− 0.45x + 0.0450.00163 = –––––––––––––––––– 4x ^{2}+ 1.6x + 0.160.00652x

^{2}+ 0.002608x + 0.0002608 = x^{2}− 0.45x + 0.0450.99348x

^{2}− 0.452608x + 0.0447392 = 0

8) Off to a quadratic equation solver:

Both roots are positive, but only one (the 0.145 value) gives valid results when plugged back into the equilibrium concentrations:[HI] ---> 0.400 + (2) (0.145) = 0.690 M

[H_{2}] ---> 0.300 − 0.145 = 0.155 M

[I_{2}] ---> 0.150 − 0.145 = 0.005 M

9) Let's plug back into the equilibrium constant expression to check:

(0.155) (0.005) K _{c}=–––––––––––– (0.69) ^{2}K

_{c}= 0.0016278Looks good to me!

**Example #10:** At a particular temperature, K_{c} = 2.0 x 10¯^{6} for the reaction:

2CO_{2}(g) ⇌ 2CO(g) + O_{2}(g)

If 2.0 mol CO_{2} is initially placed into a 5.0 L vessel, calculate the equilibrium concentrations of all species.

**Solution:**

1) An ICEbox is called for:

[CO _{2}][CO] [O _{2}]Initial 0.40 0 0 Change −2x +2x +x Equilibrium 0.40 − x 2x x The 0.40 M comes from 2.0 mol / 5.0 L.

2) Write the equilibrium constant and put values in:

[CO] ^{2}[O_{2}]K _{c}=–––––––––– [CO _{2}]^{2}

(2x) ^{2}(x)2.0 x 10¯ ^{6}=––––––––––– (0.40 − 2x) ^{2}

3) Here comes an important point: we can neglect the '2x' that is in the denominator. This is because the K_{c} is very small, which means that only a small amount of product is made. This avoids having to use a cubic equation.

4) We now continue our solution:

(2x) ^{2}(x)2.0 x 10¯ ^{6}=––––––––––– (0.40) ^{2}3.2 x 10¯

^{7}= 4x^{3}x

^{3}= 8.0 x 10¯^{8}x = 4.3 x 10¯

^{3}M

5) The three equilibrium concentrations:

[O_{2}] = x = 4.3 x 10¯^{3}M

[CO] = 2x = 8.6 x 10¯^{3}M

[CO_{2}] = 0.40 − 2x = 0.39 M

6) Let's see if neglecting the 2x was valid. To do this, we determine if the value we calculated for 2x is less than 5% of the original concentration, the 0.40.

[8.6 x 10¯^{3}/ 0.40] * 100 = 2.2%The reason for the 5% has to do with the fact that measuring equilibrium constants in the laboratory is actually quite hard. That means many equilibrium constants already have a healthy amount of error built in. That healthy amount of error has, by general practice, been set at 5%.

There are two questions asked in the bonus example. I only answered the first one and may someday address the percent yield. Also, this question was answered on Yahoo Answers in 2010. This search will give you the YA link. It's the third one down. Take a look at the first two answers, both of which give a different value than mine.

**Bonus Example Part I:** CH_{4}(g) + CO_{2}(g) ⇌ 2CO(g) + 2H_{2}(g); K_{p} = 450. at 825 K.

An 85.0 L reaction container initially contains 22.3 kg of CH_{4} and 55.4 kg of CO_{2} at 825 K.

(a) Assuming ideal gas behavior, calculate the mass of H_{2}present in the reaction mixture at equilibrium.

(b) What is the percent yield of the reaction under these conditions?

**Solution using partial pressures and K _{p}:**

1) Calculate the partial pressures of methane and carbon dioxide:

moles CH_{4}---> 22300 g / 16.0426 g/mol = 1390.05 molinitial pressure CH

_{4}:PV = nRT(P) (85.0 L) = (1390.05 mol) (0.08206 L atm / mol K) (825 K)

P

_{CH4}= 1107.126 atmmoles CO

_{2}---> 55400 g / 44.009 g/mol = 1258.83 molinitial pressure CO

_{2}:PV = nRT(P) (85.0 L) = (1258.83 mol) (0.08206 L atm / mol K) (825 K)

P

_{CO2}= 1002.614 atm

2) Create an ICEbox:

P _{CH4}P _{CO2}P _{CO}P _{H2}Initial 1107.126 1002.614 0 0 Change −x −x +2x +2x Equilibrium 1107.126 − x 1002.614 − x 2x 2x

3) Write the K_{p} expression and substitute values:

(P _{CO})^{2}(P_{H2})^{2}K _{p}=––––––––––– (P _{CH4}) (P_{CO2})

(2x) ^{2}(2x)^{2}450. = ––––––––––––––––– (1107 − x) (1003 − x)

4) Let's do the algebra leading to a quartic equation:

16x ^{4}450. = –––––––––––––––––– 1110321 − 2110x + x ^{2}16x

^{4}= 499644450 − 949500x + 450x^{2}16x

^{4}− 450x^{2}+ 949500x − 499644450 = 0

5) A quartic equation solver to the rescue:

Four roots were generated, two of which are imaginary, one is negative and the last one is positive. The positive root is 72.146. That's our value for x.

6) The pressure of hydrogen gas at equilibrium was given as '2x:'

(2) (72.146 atm) = 144.292 atmUse PV = nRT to get moles:

(144.292 atm) (85.0 L) = (n) (0.08206 L atm / mol K) (825 K)

n = 181.1656 mol

(181.1656 mol) (2.016 g/mol) = 365 g (to three sig figs)

**Bonus Example Part II:** CH_{4}(g) + CO_{2}(g) ⇌ 2CO(g) + 2H_{2}(g); K_{p} = 450. at 825 K.

An 85.0 L reaction container initially contains 22.3 kg of CH_{4} and 55.4 kg of CO_{2} at 825 K.

(a) Assuming ideal gas behavior, calculate the mass of H_{2}present in the reaction mixture at equilibrium.

(b) What is the percent yield of the reaction under these conditions?

**Solution using K _{c} and molarities:**

1) Determine initial molarities:

methane:(M) (85.0 L) = 22300 g / 16.0426 g/molM

_{CH4}= 16.3535 mol/Lcarbon dioxide:

(M) (85.0 L) = 55400 g / 44.009 g/molM

_{CO2}= 14.8098 mol/L

2) Convert K_{p} to K_{c}

The conversion is usually shown as:K_{p}= K_{c}(RT)^{Δn}where n = total moles of gas on the product side minus total moles of gas on the reactant side

Rearranging:

K _{p}K _{c}= –––––– (RT) ^{Δn}Solving:

450. K _{c}= –––––––––––––– (0.08206 x 825) ^{2}K

_{c}= 0.098184

3) Set up an ICEbox:

[CH _{4}][CO _{2}][CO] [H _{2}]Initial 16.3535 14.8098 0 0 Change −x −x +2x +2x Equilibrium 16.3535 − x 14.8098 − x 2x 2x

3) Write the K_{c} expression and substitute values:

[CO] ^{2}[H_{2}]^{2}K _{c}=––––––––––– [CH _{4}] [CO_{2}]

(2x) ^{2}(2x)^{2}0.09818 = –––––––––––––––––– (16.35 − x) (14.81 − x)

4) Let's do the algebra leading to a quartic equation:

16x ^{4}0.09818 = ––––––––––––––––––– 242.1435 − 31.16x + x ^{2}16x

^{4}= 23.77365 − 3.0593x + 0.09818x^{2}16x

^{4}− 0.09818x^{2}+ 3.0593x − 23.77365 = 0

5) A quartic equation solver to the rescue:

Four roots were generated, two of which are imaginary. One root is negative and the fourth root (a positive value) is our answer for 'x.' It is 1.066 (rounded off).

6) Determine moles of H_{2}, then grams:

(1.066 mol/L) (2) = 2.132 mol/L(2.132 mol/L) (85.0 L) = 181.22 mol

(181.22 mol) (2.016 g/mol) = 365 g (to three sig figs)