Mass-Mass Problems

#11 - 25

**Problem #11:** NH_{3} chemically reacts with oxygen gas to produce nitric oxide and water. What mass of nitric oxide is produced by the reaction of 6.40 g of oxygen gas?

**Solution:**

1) Write the balanced chemical equation:

4NH_{3}(g) + 5O_{2}(g) ---> 4NO(g) + 6H_{2}O(g)

2) The solution, done in the style of dimensional analysis:

1 mol O _{2}4 mol NO 30.006 g NO 6.40 g O _{2}x––––––––– x ––––––– x –––––––––– = 4.80 g NO 31.998 g O _{2}5 mol O _{2}1 mol NO written in one line, it looks like this:

6.40 g O

_{2}x (1 mol O_{2}/ 31.998 g O_{2}) x (4 mol NO / 5 mol O_{2}) x (30.006 g NO / 1 mol NO) = 4.80 g NO

3) Written in individual steps, the solution looks like this:

(a) convert grams of O_{2}to moles:6.40 g / 31.998 g/mol = 0.2000125 mol O_{2}(b) use a ratio and proportion involving O

_{2}and NO:

4 mol NO x ––––––– = ––––––– 5 mol O _{2}0.2000125 mol O _{2}x = 0.16001 mol NO

(c) convert moles of NO to grams:

(0.16001 mol) (30.006 g/mol) = 4.80 g NO

**Problem #12:** How many grams of magnesium nitrate can be formed from 20.00 g of oxygen gas?

**Solution:**

1) Let us write a balanced chemical equation:

Mg + N_{2}+ 3O_{2}---> Mg(NO_{3})_{2}The key point will be the 3:1 ratio between O

_{2}and Mg(NO_{3})_{2}.By the way, the above chemical reaction does not occur in nature, but the coefficients do accurately reflect how much Mg, N

_{2}and O_{2}are needed to make magnesium nitrate.

2) Determine how many moles are in the 20.00 g of O_{2}:

20.00 g / 31.9988 g/mol = 0.62502344 mol

3) We will now use the 3:1 ratio:

3 0.62502344 mol ––– = ––––––––––––– 1 x x = 0.208341 mol of Mg(NO

_{3})_{2}

4) Determine the mass of magnesium nitrate required:

(148.313 g/mol) (0.208341 mol) = 30.90 g (to four sig figs)

**Problem #13:** Water decomposes. How many moles of oxygen can be produced from 2.40 g of water at STP?

**Solution:**

1) Write a balanced chemical equation:

2H_{2}O ---> 2H_{2}+ O_{2}

2) Determine moles of water:

2.40 g / 18.015 g/mol = 0.13322 mol

3) The water to oxygen molar ratio is 2:1. Determine moles of oxygen produced:

2 is to 1 as 0.13322 mol is to xx = 0.0666 mol (to three sig figs)

Note that, since this is a mass-based problem, there is no need to use STP anywhere in the calculation.

**Problem #14:** Given this equation:

2Al(s) + 3CuSO_{4}(aq) ---> 3Cu(s) + Al_{2}(SO_{4})_{3}(aq).(a) From 25.80 g of aluminum, how many grams of copper are produced?

(b) How many moles of aluminum sulfate are produced?

**Solution to (a):**

1) Determine moles of Al:

25.80 g / 26.98 g/mol = 0.956264 mol (I kept a few gaurd digits)

2) Use Al:Cu molar ratio to determine moles of Cu produced:

2 is to 3 as 0.956264 mol is to xx = 1.434396 mol of Cu

3. Determine grams of copper:

1.434396 mol times 63.546 g/mol = 91.15 g (to four sig figs)

**Solution to (b):**

1) Use Al:Al_{2}(SO_{4})_{3} molar ratio:

2 is to 1 as 0.956264 mol is to xx = 0.478132 mol of Al

_{2}(SO_{4})_{3}

2) Determine grams of aluminum sulfate:

0.478132 mol times 342.147 g/mol = 163.6 g (to four sig figs)

OF_{2}(g) + H_{2}O(ℓ) ---> 2HF(aq) + O_{2}(g)

**Solution:**

1) Determine moles of OF_{2}:

108.0 g / 53.995 g/mol = 2.000 mol

2) Use OF_{2}:H_{2}O molar ration to determine moles of H_{2}O:

1 is to 1 as 2.000 mol is to xx = 2.000 mole of H

_{2}O

3) Determine grams of H_{2}O:

2.000 mol times 31.9988 g/mol = 64.00 g (to four sig figs)

**Problem #16:** Based on this balanced equation:

10Li + N_{2}F_{4}---> 4LiF + 2Li_{3}N

(a) Calculate the formula units of LiF formed when 570 atoms of Li are reacted.

(b) Calculate the formula units of LiF formed, if 570 atoms of N are used in the reaction along with sufficient Li.

**Solution to a:**

Use the Li:LiF molar ratio to determine formula units of LiF produced:

10 is to 4 as 570 atoms is to xx = 228 formula units of LiF

Note: if you wished to reduce the 10:4 ratio to 5:2 before calculating, the 5:2 ratio would lead to the correct answer.

Please be aware that discussing NUMBERS of atoms or formula units is analogous to using MOLES. Remember that using moles is simply a short-hand for discussing how many atoms or formula units are present. (Reminder: one mole contains 6.022 x 10^{23} of the entities -- be they atoms, molecules or formula units -- under discussion.)

**Solution to b:**

1) In every one molecule of N_{2}F_{4}, there are two atoms of N. We need to know how many molecules of N_{2}F_{4} are present:

570 atoms of N divided by 2 atoms of N per one molecule of N_{2}F_{4}= 285 molecules of N_{2}F_{4}

2) Use the N_{2}F_{4} to LiF ratio to determine formula units of LiF

1 is to 4 as 285 is to xx = 1140 formula units of LiF

Notice the extra step. The N_{2}F_{4} reacts as a molecular unit, not individual atoms of N. Hence, I needed to determine how many molecules of N_{2}F_{4} were composed of 570 atoms of N.

I did not have to do this in (a) because the Li reacted on the basis of individual atoms of Li. If the formula had been Li_{2}, I would have divided 570 by 2. if it had been Li_{4}, I would have divided by 4.

**Problem #17:** 46.0 g of an alkai metal was reacted with water to form the aqueous metal hydroxide along with 1.19 g of hydrogen gas. Which alkai metal was used?

**Solution:**

1) Let M be the alkali metal. The chemical reaction is this:

2M + 2H_{2}O ---> 2MOH + H_{2}

2) Determine moles of H_{2} produced:

1.19 g / 2.016 g/mol = 0.59028 mol

3) Moles M required:

Molar ratio between M and H_{2}is 2:12 is to 1 as x is to 0.59028

x = 1.18056 mol

4) Determine atomic weight of M:

46.0 g / 1.18056 mol = 39.0 g/molM is potassium.

**Problem #18:** Mg + 2HCl ---> MgCl_{2} + H_{2}

(a) How many grams of magnesium (Mg) are needed to produce 100.0 grams of hydrogen (H_{2})?

(b) How many grams of hydrogen chloride (HCl) is needed to produce 200.0 g of hydrogen (H_{2})?

(c) If 500. g of magnesium chloride (MgCl_{2}) are produced in the above reaction, how many grams of hydrogen (H_{2}) would be produced?

**Solution to (a):**

1 mol H _{2}1 mol Mg 24.305 g Mg 100.0 g H _{2}x––––––––– x ––––––– x –––––––––– = 1206 g Mg (to four sig figs) 2.016 g H _{2}1 mol H _{2}1 mol Mg

**Solution to (b):**

1 mol H _{2}2 mol HCl 36.4609 g HCl 200.0 g H _{2}x––––––––– x ––––––––––– x –––––––––––– = 7234 g HCl 2.016 g H _{2}1 mol MgCl _{2}2 mol HCl

**Solution to (c):**

1 mol MgCl _{2}1 mol H _{2}2.016 g H _{2}500.0 g MgCl _{2}x–––––––––––– x –––––––––– x ––––––––– = 10.59 g H _{2}95.211 g MgCl _{2}1 mol MgCl _{2}1 mol H _{2}

**Problem #19:** P_{4}(s) + 5O_{2}(g) ---> P_{4}O_{10}(g)

(a) How many grams of phosphorus(V) oxide (P_{4}O_{10}) are produced if you burn 50.0 grams of phosphorus with sufficient oxygen (O_{2})?

(b) How many grams of oxygen would be needed in part (a)?

(c) If 400. grams of phosphorus(V) oxide (P_{4}O_{10}) is needed for another experiment, how much phosphorus would have to be burned?

**Solution to (a):**

1) Determine moles of phosphorus that burned:

50.0 g / 123.896 g/mol = 0.403564 mol

2) Determine moles of P_{4}O_{10} produced:

1 1000 g ––––––– = ––––––– 1 x x = 0.403564 mol (of P

_{4}O_{10}, NOT P_{4})

3) Determine grams of P_{4}O_{10} produced:

(0.403564 mol) (283.886 g/mol) = 114 g (to three sig figs)

**Solution to (b):**

1 mol P _{4}5 mol O _{2}32.00 g O _{2}50.0 g P _{4}x–––––––––– x –––––––– x ––––––––– = 64.57 g H _{2}123.896 g P _{4}1 mol P _{4}1 mol O _{2}

**Solution to (c):**

1) Determine moles of P_{4}O_{10} produced:

400. g / 283.886 g/mol = 1.4090163 mol

2) The molar ratio between P_{4} and P_{4}O_{10} is 1:1

Therefore, 1.4090163 mol of P_{4}is required.

3) Determine grams of P_{4} required:

(1.4090163 mol) (123.896 g/mol) = 174.57 gTo three sig figs, this is 174 g (remember the rule of five for rounding)

**Problem #20:** The Claus reactions, shown below, are used to generate elemental sulfur from hydrogen sulfide.

2H_{2}S + 2O_{2}--->^{1}⁄_{8}S_{8}+ SO_{2}+ 2H_{2}O

2H_{2}S + SO_{2}--->^{3}⁄_{8}S_{8}+ 2H_{2}O

What mass of sulfur can be produced from 48.0 grams of O_{2}?

**Solution (short way):**

1) Let's add the two equations together:

4H_{2}S + 2O_{2}--->^{1}⁄_{2}S_{8}+ 4H_{2}OThere is a 4:1 molar ratio between O

_{2}and S_{8}

2) Determine moles of oxygen:

48.0 g / 32.0 g/mol = 1.50 mol

3) Using the molar ratio, determine moles of sulfur produced:

4 is to 1 as 1.50 mol is to xx = 0.375 mol of S

_{8}

4) Determine mass of sulfur produced:

(0.375 mol) (256.52 g/mol) = 96.2 g

**Problem #21:** In an experiment, potassium chlorate decomposed according to the following chemical equation.

2KClO_{3}---> KCl + 3O_{2}

If the mass of potassium chlorate was 240. g, which of the following calculations can be used to determine the mass of oxygen gas formed?

(a) (240 x 2 x 32.00) ÷ (122.5 x 3) g

(b) (240 x 3 x 32.00) ÷ (122.5 x 2) g

(c) (240 x 2 x 122.5) ÷ (32.00 x 3) g

(d) (240 x 3 x 122.5) ÷ (32.00 x 2) g

**Solution:**

1) Set up using dimensional analysis:

1 mol 3 mol O _{2}32.00 g 240. g x ––––––– x ––––––– x ––––––– = 94.0 g O _{2}122.5 g 2 mol KClO _{3}1 mol

2) We examine the answer choices, looking for:

240, 3, and 32 in the numerator

122.5 and 2 in the denominatorWe find that answer choice (b) fits our needs.