Limiting Reagent Problems solved using only dimensional analysis

Return to main mass-mass stoichiometry tutorial

**Example #1:** Lithium hydroxide is used in an outer space environment to remove excess exhaled carbon dioxide from the living environment. The products of the reaction are lithium carbonate and water. If 48.0 grams of lithium hydroxide
was used in a small scale experimental device, how many grams of carbon dioxide will the device process? The balanced equation is:

2LiOH(s) + CO_{2}(g) ---> Li_{2}CO_{3}(s) + H_{2}O(ℓ)

**Solution:**

1) Write the given information in the first numerator:

48.0 g LiOH ––––––––––– x ––––––– x –––––––– x ––––––– = answer 1 Notice how I included the formula of the compound. There's a reason for that and you'll see it in just a moment.

2) The next step is to convert the grams given in the problem to moles:

48.0 g LiOH 1 mol LiOH ––––––––––– x –––––––––––– x –––––––– x ––––––– = answer 1 23.948 g LiOH An important point about DA is that you should cancel units on the diagonal. In our example, that means the 'g LiOH' on the 48.0 cancels with the 'g LiOH' on the 23.948.

You should not attempt to cancel vertically since the unit 'mol LiOH' is different from the unit "g LiOH.'

3) The molar ratio should be placed next. Remember, a molar ratio uses moles, not grams. Here is the result:

48.0 g LiOH 1 mol LiOH 1 mol CO _{2}––––––––––– x –––––––––––– x –––––––––– x ––––––– = answer 1 23.948 g LiOH 2 mol LiOH I placed the '2 mol LiOH' in the denominator because I wanted it to cancel the unit 'mol LiOH' in the numerator, the '1 mol LiOH.' Note that a new unit (the 'mol CO

_{2}') is introduced when the molar ratio is complete.

4) We know that the final answer will be in grams. We need to place the molecular weight of carbon dioxide in such a way as to cancel 'mol CO_{2}' and introduce 'g CO_{2}' Here's how we do that:

48.0 g LiOH 1 mol LiOH 1 mol CO _{2}44.009 g CO _{2}––––––––––– x –––––––––––– x –––––––––– x ––––––––––– = 44.1 g CO _{2}1 23.948 g LiOH 2 mol LiOH 1 mol CO _{2}

**Example #2:** Metallic iron reacts with oxygen gas to produce iron(III) oxide. Given that 64.0 g of oxygen gas is present, determine how many grams of iron react.

**Solution:**

1) Write the balanced chemical equation:

4Fe + 3O_{2}---> 2Fe_{2}O_{3}

2) Write the given information in the first numerator:

64.0 g O _{2}––––––––––– x ––––––– x –––––––– x ––––––– = answer 1 Sometimes the '1' in the denominator is not written. There will be situation where the '1' will go into the numerator.

3) Divide the mass by the molar mass of oxygen gas:

64.0 g O _{2}1 mol O _{2}––––––––––– x ––––––– x –––––––– x ––––––– = answer 1 32.0 g O _{2}

4) Create the molar ratio between O_{2} and Fe:

64.0 g O _{2}1 mol O _{2}––––––––––– x ––––––– x –––––––– x ––––––– = answer 1 32.0 g O _{2}3 mol O _{2}I placed the '3 mol O

_{2}' in the denominator so as to cancel with the 'mol O_{2}' unit in the numerator of the second fraction.

5) Place the second part of the molar ratio:

64.0 g O _{2}1 mol O _{2}4 mol Fe ––––––––––– x ––––––– x –––––––– x ––––––– = answer 1 32.0 g O _{2}3 mol O _{2}All the information before this step concerned oxygen. It is that this point in the calculation where we replace oxygen with iron.

6) Place the value that will change 'mol Fe' to 'g Fe'

64.0 g O _{2}1 mol O _{2}4 mol Fe 55.845 g Fe ––––––––––– x ––––––– x –––––––– x –––––––––– = 149 g Fe (to three sig figs) 1 32.0 g O _{2}3 mol O _{2}1 mol Fe When you get to the molar ratio (the 4/3 part), you can multiply by 4, then divide by 3. Or, you can notice that four-thirds is 1.3333333 . . . , so you could just multiply by 1.333333 . . . Use enough threes to really make sure that that number doesn't affect the final answer. In other words, don't use 1.3 or even 1.33, use something like 1.3333333.

By the way, the 4 and the 3 of the molar ratio do not affect sig figs in the final answer. Those numbers are exact, since they are derived from theoretical ideas. Whereas 64.0 g is a mass and those are always, in the final analysis, measured by some instrument.

**Example #3:** 95.83 g of iron is completely consumed when reacted with excess oxygen. How many grams of iron(III) oxide are produced?

**Solution:**

1) The balanced equation:

4Fe + 3O_{2}---> 2Fe_{2}O_{3}

2) Start with the information given in the problem:

95.83 g Fe ––––––––––– x ––––––– x –––––––– x ––––––– = answer This time I decided to leave the denominator blank.

3) Convert grams to moles:

95.83 g Fe 1 mol Fe ––––––––––– x ––––––– x –––––––– x ––––––– = answer 55.845 g Fe

4) Create the molar ratio:

95.83 g Fe 1 mol Fe 1 mol Fe _{2}O_{3}––––––––––– x ––––––– x –––––––––– x ––––––– = answer 55.845 g Fe 2 mol Fe The molar ratio from the coefficients is 2 to 4. I took out the common factor of 2 to give a new ratio of 1 to 2. he two ratios are, of course, equal to each other.

If you used a 2 to 4 ratio in your calculation, you did nothing wrong. This is because 2 to 4 and 1 to 2 are the same numerical value.

In my classroom teaching (now years in the past), I tended to use the unreduced ratio and then point out that you could also used a reduced ratio.

**Example #4:** Nitrogen gas and hydrogen gas react to form ammonia as follows:

N_{2}+ 3H_{2}---> 2NH_{3}

When 8.43 g of H_{2} react completely, how many grams of ammonia are formed?

**Solution:**

1) Here is the entire dimensional analysis set up:

8.43 g H _{2}1 mol H _{2}2 mol NH _{3}17.031 g NH _{3}––––––––– x ––––––––– x –––––––– x ––––––––––– = 47.8 g NH _{3}2.016 g H _{2}3 mol H _{2}1 mol NH _{3}

2) Do you see any patterns among the five DA set ups from example #1 to example #5?

**Example #5:** 108.5 g of ammonia is produced. How many grams of hydrogen gas was consumed?

**Solution:**

1) The entire DA set up:

108.5 g NH _{3}1 mol NH _{3}3 mol H _{3}2.016 g H _{2}––––––––––– x ––––––––––– x –––––––– x ––––––––– = 19.3 g H _{2}17.031 g NH _{3}2 mol NH _{3}1 mol H _{2}(a) (b) (c) (d) <--- see comments just below

2) Comments:

(a) You always start with the value given in the problem statement. This particular type of problem is called mass-mass stoichiometry, therefore the given value will always be a mass. If it was anything else (such as moles or number of molecules), it wouldn't be a mass-mass stoichiometry problem.(b) This is always the molar mass of the given mass. It will always be in the denominator.

(c) This ratio is always structured so as to cancel the unit in the numerator (the '1 mol NH

_{3}' unit in example #5)(d) This is always the molar mass of the substance that is the unknown. It will always be in the numerator.

**Example #6:** 50.00 g of Na_{3}PO_{4} and 30.00 g of NH_{4}NO_{3} are allowed to react. Given this reaction:

Na_{3}PO_{4}+ 3NH_{4}NO_{3}---> (NH_{4})_{3}PO_{4}+ 3NaNO_{3}

(a) How much of each product is produced?

(b) How much of the excess reagent remains when the reaction stops?

**Solution:**

1) We will do two calculations, one in which we act as if Na_{3}PO_{4} is the limiting reagent and one as if NH_{4}NO_{3} is the limiting reagent. In each calculation, we will pick the same product. The lesser amount of product produced will indicate the limiting reagent.

2) Assume Na_{3}PO_{4} is the limiting reagent:

50.00 g Na _{3}PO_{4}1 mol Na _{3}PO_{4}1 mol (NH _{4})_{3}PO_{4}149.0858 g (NH _{4})_{3}PO_{4}––––––––––––– x ––––––––––––– x –––––––––––––– x –––––––––––––––––– = 45.47 g (NH _{4})_{3}PO_{4}163.94 g Na _{3}PO_{4}1 mol Na _{3}PO_{4}1 mol (NH _{4})_{3}PO_{4}

3) Assume NH_{4}NO_{3} is the limiting reagent:

30.00 g NH _{4}NO_{3}1 mol NH _{4}NO_{3}1 mol (NH _{4})_{3}PO_{4}149.0858 g (NH _{4})_{3}PO_{4}––––––––––––– x ––––––––––––– x –––––––––––––– x –––––––––––––––––– = 18.62 g (NH _{4})_{3}PO_{4}80.0426 g NH _{4}NO_{3}3 mol NH _{4}NO_{3}1 mol (NH _{4})_{3}PO_{4}

4) The ammonium nitrate is the limiting reagent.

5) The calculation in step 3 just above gives the answer t how much (NH_{4})_{3}PO_{4}
is produced. A new calculation must be performed to determine how much NaNO_{3} was produced.

30.00 g NH _{4}NO_{3}1 mol NH _{4}NO_{3}3 mol NaNO _{3}84.994 g NaNO _{3}–––––––––––––– x –––––––––––––––– x –––––––––––– x –––––––––––––– = 31.86 g NaNO _{3}80.0426 g NH _{4}NO_{3}3 mol NH _{4}NO_{3}1 mol NaNO _{3}

6) We know that there are 80.00 g of reactants and we know that 50.48 g of products were produced. Therefore, 29.52 g of sodium phosphate remains. It must be this way in order to satisfy the Law of Conservation of Mass.

7) Here is the stoichiometric calculation that gives us the amount used up of the excess reactant. We then do a subtraction to get the amount of the excess reagent that remains after the reaction ceases.

30.00 g NH _{4}NO_{3}1 mol NH _{4}NO_{3}1 mol Na _{3}PO_{4}163.94 g Na _{3}PO_{4}––––––––––––– x ––––––––––––––– x –––––––––––– x –––––––––––––– = 20.48 g Na _{3}PO_{4}80.0426 g NH _{4}NO_{3}3 mol NH _{4}NO_{3}1 mol Na _{3}PO_{4}

8) 20.48 g is the amount used up. The last step is to subtract the used amount for the amount present a the start.

50.00 − 20.48 = 29.52 g remain