Stoichiometry Practice

Every question here is three steps wearing different clothes:

$mass given \to moles \to moles \to mass wanted$

The first arrow is $n = \frac{m}{M}$ . The middle arrow — the only one that is actually chemistry — is the coefficient ratio from the balanced equation. The last arrow is $m = n M$ again.

Balance the equation before you write anything else. A perfect calculation on an unbalanced equation is worth nothing, and there is no partial credit route back from it.

Assume enough of everything else is present unless a question says otherwise. Molar masses to two decimal places; round once, at the end.

1. For $2 H X_{2} + O X_{2} 2 H X_{2} O$ , how many moles of water are produced from 3.0 mol of hydrogen? How many moles of oxygen are consumed?

Answer 1

No molar masses needed — you were given moles and asked for moles, so only the middle arrow is in play.

Water. The ratio of hydrogen to water in the equation is 2 to 2, which is 1 to 1:

$n (H X_{2} O) = 3.0 mol \times \frac{2}{2} = 3.0 mol$

Oxygen. The ratio of hydrogen to oxygen is 2 to 1:

$n (O X_{2}) = 3.0 mol \times \frac{1}{2} = 1.5 mol$

Write the ratio as a fraction with the substance you want on top and the substance you have underneath, every single time. Done that way the arithmetic tells you when you have inverted it, because the answer comes out obviously the wrong size.

2. In the reaction $N X_{2} + 3 H X_{2} 2 NH X_{3}$ , what mass of ammonia is produced from 28.0 g of nitrogen?

Answer 2

$M (N X_{2}) = 2 (14.01) = 28.02 g/mol$ and $M (NH X_{3}) = 14.01 + 3 (1.01) = 17.04 g/mol$

$n (N X_{2}) n (NH X_{3}) m (NH X_{3}) = \frac{28.0 g}{28.02 g/mol} = 0.99929 mol = 0.99929 \times \frac{2}{1} = 1.9986 mol = (1.9986) (17.04) = 34.06 g$

34.1 g of ammonia, to three significant figures.

Two checks worth doing. The mass went up, which it should — two nitrogen atoms have become two ammonia molecules that each picked up three hydrogens. And 28.0 g of nitrogen is almost exactly one mole, so the answer should be close to twice the molar mass of ammonia, which is 34.08. It is.

3. Calcium carbonate decomposes on strong heating: $CaCO X_{3} CaO + CO X_{2}$ . Starting from 50.0 g of calcium carbonate, find the mass of each product.

Answer 3

$M (CaCO X_{3}) = 40.08 + 12.01 + 3 (16.00) = 100.09 g/mol$

$n (CaCO X_{3}) = \frac{50.0}{100.09} = 0.49955 mol$

Both coefficient ratios are 1 to 1, so both products form in that same amount.

Carbon dioxide: $M = 44.01$ g/mol, so $m = (0.49955) (44.01) = 21.99 g$ — 22.0 g.

Calcium oxide: $M = 40.08 + 16.00 = 56.08$ g/mol, so $m = (0.49955) (56.08) = 28.01 g$ — 28.0 g.

Now the free check that costs five seconds and catches almost everything: $22.0 + 28.0 = 50.0$ g, which is exactly what you started with. Mass is conserved, so the product masses must add to the reactant mass, and if they do not you have made an arithmetic error somewhere upstream.

This reaction runs in every lime kiln in the country, and the carbon dioxide in that answer is not a by-product of the fuel — it comes out of the rock itself. That is a fact worth carrying into Chemistry at Industrial Scale.

4. Iron reacts with copper(II) sulfate solution: $Fe + CuSO X_{4} FeSO X_{4} + Cu$ . What mass of copper should 2.00 g of iron produce?

Answer 4

$n (Fe) n (Cu) m (Cu) = \frac{2.00 g}{55.85 g/mol} = 0.035810 mol = 0.035810 \times \frac{1}{1} = 0.035810 mol = (0.035810) (63.55) = 2.2757 g$

2.28 g of copper.

The mass went up even though the mole ratio is 1 to 1, because a copper atom is heavier than an iron atom. This is worth holding on to: a one-to-one mole ratio is not a one-to-one mass ratio, and the only reactions where mass in equals mass out for a single pair are the ones where the two molar masses happen to match.

This is also a theoretical yield. What you would actually recover from a beaker is less, for the reasons set out in Percentage Yield of a Precipitate.

5. Potassium chlorate decomposes on heating: $2 KClO X_{3} 2 KCl + 3 O X_{2}$ . What mass of oxygen is produced from 12.25 g of potassium chlorate?

Answer 5

$M (KClO X_{3}) = 39.10 + 35.45 + 3 (16.00) = 122.55 g/mol$

$n (KClO X_{3}) n (O X_{2}) m (O X_{2}) = \frac{12.25 g}{122.55 g/mol} = 0.099959 mol = 0.099959 \times \frac{3}{2} = 0.14994 mol = (0.14994) (32.00) = 4.7980 g$

4.798 g of oxygen, to four significant figures — the mass was given to four and the molar masses support it.

The $\frac{3}{2}$ is the whole question. Every other step is bookkeeping, and a student who writes $n (O X_{2}) = 0.0999$ mol has done all the arithmetic correctly and answered a different question. Circle the coefficients in the equation before you start; it takes two seconds and it is where the marks are.

6. Propane burns according to $C X_{3} H X_{8} + 5 O X_{2} 3 CO X_{2} + 4 H X_{2} O$ . For 10.0 g of propane, find the mass of oxygen required and the mass of each product.

Answer 6

$M (C X_{3} H X_{8}) = 3 (12.01) + 8 (1.01) = 44.11 g/mol$

$n (C X_{3} H X_{8}) = \frac{10.0}{44.11} = 0.22671 mol$

Now one coefficient ratio per substance, all from the same starting amount:

$m (O X_{2}) m (CO X_{2}) m (H X_{2} O) = 0.22671 \times \frac{5}{1} \times 32.00 = 36.27 g = 0.22671 \times \frac{3}{1} \times 44.01 = 29.93 g = 0.22671 \times \frac{4}{1} \times 18.02 = 16.34 g$

36.3 g of oxygen, 29.9 g of carbon dioxide, and 16.3 g of water.

The conservation check: in, $10.0 + 36.27 = 46.27$ g; out, $29.93 + 16.34 = 46.27$ g. They match exactly on the unrounded values.

Add up the rounded answers instead and you get 46.3 g in against 46.2 g out. That gap of 0.1 g is not a physics problem and not a mistake — it is what rounding three separate numbers to three figures does. It is also the clearest possible demonstration of why you round once, at the end, and why a conservation check should be run on the full-precision values.

One more thing hiding in this answer: burning 10.0 g of propane requires more than three and a half times its own mass in oxygen, all of it taken from the air. Nearly five times the fuel’s mass leaves as exhaust. Anyone who has wondered how a car burning fifty kilograms of fuel emits more than fifty kilograms of carbon dioxide has just been shown the reason.

7. In $2 H X_{2} + O X_{2} 2 H X_{2} O$ , does 2 g of hydrogen react with 1 g of oxygen? Say exactly what the coefficients do and do not tell you, and work out the actual mass ratio.

Answer 7

No. The coefficients count particles, never grams. Reading them as masses is the single most expensive misreading available in this unit, because it produces answers that look reasonable.

What the equation says is: for every two hydrogen molecules consumed, one oxygen molecule is consumed and two water molecules are made. To get masses you have to go through the molar masses.

$m (2 mol H_{2}) m (1 mol O_{2}) = 2 \times 2.02 = 4.04 g = 1 \times 32.00 = 32.00 g$

So the mass ratio is $4.04 : 32.00$ , which is about 1 to 7.92. Hydrogen and oxygen react in a 2 to 1 ratio by particles and roughly a 1 to 8 ratio by mass, and both statements describe the same reaction.

Check the products while you are here: two moles of water is $2 \times 18.02 = 36.04$ g, and $4.04 + 32.00 = 36.04$ g. Conserved, as it must be.

The useful summary: coefficients are a recipe in particles, and your balance is a scale in grams. The mole is the only thing that connects them, which is why The Mole sits where it does in this course.

8. Three claims about question 3. Say what is wrong with each. (a) ” $n (CaCO X_{3}) = 50.0 \times 100.09 = 5004.5$ mol.” (b) “The mole ratio is 1 to 1, so 50.0 g of calcium carbonate gives 50.0 g of carbon dioxide.” (c) “Mass is conserved, so the mass of carbon dioxide must equal the mass of the calcium carbonate.”

Answer 8

(a) Multiplied instead of divided, and two independent checks catch it instantly.

The units: grams multiplied by grams per mole gives grams squared per mole, which is not a quantity. The correct operation is $n = \frac{m}{M} = \frac{50.0}{100.09} = 0.500$ mol.

The magnitude: 50.0 g of a solid you could hold in one hand is not five thousand moles of anything. One mole of most solids is a spoonful to a small pile, so an answer in the thousands should stop you before you write the next line. Get into the habit of asking “is this the right size” — it is worth more than any formula on the page.

(b) Confusing a mole ratio with a mass ratio. The ratio 1 to 1 means one mole of carbon dioxide per mole of calcium carbonate, and those two moles have very different masses: 44.01 g against 100.09 g. The correct answer is 22.0 g, as worked above. This is the same error as question 7 in a different disguise, and it is the error that most often survives all the way to a final answer without anything looking wrong.

(c) The principle is right and it has been applied to the wrong set of substances. Mass is conserved across the whole reaction, not between one reactant and one product. There are two products here, and the 50.0 g is shared between them: 22.0 g leaves as carbon dioxide and 28.0 g stays behind as calcium oxide. Conservation says $50.0 = 22.0 + 28.0$ , which it does.

The version of this claim that would be true: the total mass of all products equals the total mass of all reactants. Dropping the word “total” turns a law into a mistake.

Reference: Stoichiometry and The Mole. When one reactant runs out before the others: Limiting Reagent Practice.

Curriculum connection

D2.5
calculate the corresponding mass, or quantity in moles or molecules, for any given reactant or product in a balanced chemical equation as well as for any other reactant or product in the chemical reaction [AI]
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D3.4
explain the quantitative relationships expressed in a balanced chemical equation, using appropriate units of measure (e.g., moles, grams, atoms, ions, molecules)
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Stoichiometry Practice

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