How do you solve a gas law stoichiometry problem?

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How do you solve a gas law stoichiometry problem?

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Answer 1 · 2014-01-03T13:19:54

The easiest way is to remember that in order to use stoichiometry, you need to know the moles of the two substances concerned.

Explanation:

We can use the gas laws to help us to determine the effect of temperature, pressure, and volume on the number of moles of a gas.

The central requirement of any stoichiometry problem is to convert moles of $A$ $"A"$ to moles of $B$ $"B"$ .

If $A$ $"A"$ and/or $B$ $"B"$ are solids or liquids, you use the mass and molar mass to get moles.

If $A$ $"A"$ and/or $B$ $"B"$ are gases, you use the Ideal Gas Law to get moles.

Here's a flow chart to help you through the process.

Flow Chart

(From homepage.usask.ca)

EXAMPLE 1

What volume of oxygen at STP is produced when 10.0 g of potassium chlorate decomposes to form potassium chloride and oxygen?

Solution

First, you need the balanced chemical equation for the reaction.
You then use the molar mass to convert grams of potassium chlorate to moles of potassium chlorate.
Next, the central part of the problem is to get the molar ratio between potassium chlorate and oxygen. This gives you the moles of oxygen.
Finally, you use the molar volume to convert moles to litres.

Let’s see how this works.

Step 1. Write the balanced equation.

${2KClO}_{3} \to 2KCl + {3O}_{2}$ $"2KClO"_3 → "2KCl" + "3O"_2$

Step 2. Calculate the moles of ${KClO}_{3}$ $"KClO"_3$ .

$10.0 g KClO₃ \times \frac{{1 mol KClO}_{3}}{122.6 g KClO₃} = {0.08156 mol KClO}_{3}$ $10.0 cancel("g KClO₃") × ("1 mol KClO"_3)/(122.6 cancel("g KClO₃")) = "0.08156 mol KClO"_3$

Step 3. Calculate the moles of $O_{2}$ $"O"_2$ .

The balanced equation tells us that 2 mol ${KClO}_{3}$ $"KClO"_3$ give 3 mol $O_{2}$ $"O"_2$ . Therefore,

$0.08156 mol KClO₃ \times \frac{{3 mol O}_{2}}{2 mol KClO₃} = {0.1223 mol O}_{2}$ $0.08156 cancel("mol KClO₃") × ("3 mol O"_2)/(2 cancel("mol KClO₃")) = "0.1223 mol O"_2$

Step 4. Convert moles of $O_{2}$ $"O"_2$ to litres of $O_{2}$ $"O"_2$ .

Since 1997, STP has been defined as 0 °C and 100 kPa.

The molar volume of an ideal gas at STP is 22.711 L.

At STP, we use the relation 22.711 L = 1 mol. Therefore

$0.1223 mol O₂ \times \frac{{22.711 L O}_{2}}{1 mol O₂} = {2.78 L O}_{2}$ $0.1223 cancel("mol O₂") × ("22.711 L O"_2)/(1 cancel("mol O₂")) = "2.78 L O"_2$

Notice how we always write the conversion factors so that the units cancel to give the desired units for the answer.

If the question asks you to find the volume of gas at some other temperature or pressure, you can use the Ideal Gas Law,

$P V = n R T$ $PV = nRT$ .

Suppose the question had asked for the volume at 1.05 atm and 25 °C (298 K). You would write

$V = \frac{n R T}{P} = \frac{0.122 mol \times 0.082 06 L\cdot atm\cdotK⁻¹mol⁻¹ \times 298 K}{1.05 atm} = 2.85 L$ $V = (nRT)/P = (0.122 cancel("mol") × "0.082 06 L·"cancel("atm·K⁻¹mol⁻¹") × 298 cancel("K"))/(1.05 cancel("atm")) = "2.85 L"$

EXAMPLE 2

What mass of potassium chlorate is required to produce 3.00 L of oxygen at STP?

Solution

First, you need the balanced chemical equation for the reaction.
Next, you use the molar volume to convert litres to moles.
The central part of the problem is to get the molar ratio between potassium chlorate and oxygen. This gives you the moles of potassium chlorate.
You then use the molar mass to convert moles of potassium chlorate to grams of potassium chlorate.

Step 1. The balanced equation is

${2KClO}_{3} \to 2KCl + {3O}_{2}$ $"2KClO"_3 → "2KCl" + "3O"_2$

Step 2. Convert litres at STP to moles.

$3.00 L O₂ \times \frac{{1 mol O}_{2}}{22.711 L O₂} = {0.1321 mol O}_{2}$ $3.00 cancel("L O₂") × ("1 mol O"_2)/(22.711 cancel("L O₂")) = "0.1321 mol O"_2$

Step 3. Convert moles of $O_{2}$ $"O"_2$ to moles of ${KClO}_{3}$ $"KClO"_3$ .

$0.1321 mol O₂ \times \frac{{2 mol KClO}_{3}}{3 mol O₂} = {0.088 06 mol KClO}_{3}$ $0.1321 cancel("mol O₂") × ("2 mol KClO"_3)/(3 cancel("mol O₂")) = "0.088 06 mol KClO"_3$

Step 4. Calculate the moles of ${KClO}_{3}$ $"KClO"_3$ .

$0.08806 mol KClO₃ \times \frac{{122.6 g KClO}_{3}}{1 mol KClO₃} = {10.8 g KClO}_{3}$ $0.08806 cancel("mol KClO₃") × ("122.6 g KClO"_3)/(1 cancel("mol KClO₃")) = "10.8 g KClO"_3$

If the question asks you to find the volume of gas at some other temperature or pressure, you can use the Ideal Gas Law, $P V = n R T$ $PV = nRT$ .

Suppose the question gave you the volume at 1.05 atm and 25°C (298 K). You would write

$n = \frac{P V}{R T} = \frac{1.05 atm \times 3.00 L}{0.082 06 L\cdotatm\cdotK⁻¹ {mol}^{- 1} \times 298 K} = {0.129 mol O}_{2}$ $n = (PV)/(RT) = (1.05 cancel("atm") × 3.00 cancel("L"))/(0.082 06 cancel("L·atm·K⁻¹")"mol"^-1 × 298 cancel("K")) = "0.129 mol O"_2$

Now that you have the moles of $O_{2}$ $"O"_2$ , you can continue with steps 3 and 4 above.

EXAMPLE 3

Ethylene gas burns in air according to the following equation.

$C_{2} H_{4} (g) + {3O}_{2} (g) \to {2CO}_{2} (g) + {2H}_{2} O(l)$ $"C"_2"H"_4"(g)" + "3O"_2"(g)" → "2CO"_2"(g)" + "2H"_2"O(l) "$

If 13.8 L of $C_{2} H_{4}$ $"C"_2"H"_4$ at 21°C and 1.083 atm burns completely in oxygen, calculate the volume of ${CO}_{2}$ $"CO"_2$ produced, assuming the ${CO}_{2}$ $"CO"_2$ is measured at 44°C and 0.989 atm.

Solution

This one requires a little more work, because you have to use the Ideal Gas Law at the beginning and at the end.

You already have the balanced chemical equation, so your first task is to use the Ideal Gas Law to calculate the moles of $C_{2} H_{4}$ $"C"_2"H"_4$ .
The central part of the problem is to get the molar ratio between ${CO}_{2}$ $"CO"_2$ and $C_{2} H_{4}$ $"C"_2"H"_4$ . This gives you the moles of ${CO}_{2}$ $"CO"_2$ .
You then use the Ideal Gas Law to convert moles of ${CO}_{2}$ $"CO"_2$ to litres of ${CO}_{2}$ $"CO"_2$ under the new conditions.

Let’s see how this works. The balanced equation is

$C_{2} H_{4} (g) + {3O}_{2} (g) \to {2CO}_{2} (g) + {2H}_{2} O(l)$ $"C"_2"H"_4"(g) "+ "3O"_2"(g)" → "2CO"_2"(g)" + "2H"_2"O(l)"$

Step 1. Calculate the moles of $C_{2} H_{4}$ $"C"_2"H"_4$ .

$P V = n R T$ $PV = nRT$

$n = \frac{P V}{R T} = \frac{1.083 atm \times 13.80 L}{0.082 06 L\cdotatm\cdotK⁻¹ {mol}^{- 1} \times 294 K} = {0.619 mol C}_{2} H_{4}$ $n = (PV)/(RT) = (1.083 cancel("atm") × 13.80 cancel("L"))/(0.082 06 cancel("L·atm·K⁻¹")"mol"^-1 × 294 cancel("K")) = "0.619 mol C"_2"H"_4$

Step 2. Calculate the moles of ${CO}_{2}$ $"CO"_2$

$0.619 mol C₂H₄ \times \frac{{2 mol CO}_{2}}{1 mol C₂H₄} = {1.24 mol CO}_{2}$ $0.619 cancel("mol C₂H₄") × ("2 mol CO"_2)/(1 cancel("mol C₂H₄")) = "1.24 mol CO"_2$

Step 3. Calculate the new volume.

$P V = n R T$ $PV = nRT$

$V = \frac{n R T}{P} = \frac{1.24 mol \times 0.082 06 L\cdot atm\cdotK⁻¹mol⁻¹ \times 317 K}{0.989 atm} = 32.6 L$ $V = (nRT)/P = (1.24 cancel("mol") × "0.082 06 L·"cancel("atm·K⁻¹mol⁻¹") × 317 cancel("K"))/(0.989 cancel("atm")) = "32.6 L"$

Here's a great video showint the relation between stoichiometry and thr Ideal Gas Law.

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How do you solve a gas law stoichiometry problem?

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