Q.5.52 - An Introduction to General, Organic, and Biological Chemistry

Question

Use the molar volume to calculate each of the following at STP:

a. the number of moles of CO₂ in 4.00 L of CO₂  gas.

b. the volume in liters, occupied by 0.420 moles of He gas.

c. the volume, in liters, occupied by 6.40 g of O₂  gas.

d. the number of grams of Ne contained in 11.2 L of Ne gas.

Step-by-Step Solution

Answer

a. The number of moles is 0.178 moles.

b. The volume is 9.408 L.

c. The volume is  4.48 L.

d. The number of grams is 10.05 grams.

Step 1: Part (a) Given Information

The volume of CO₂ is 4.00 L.

Step 2: Part (a) Explanation

Avogadro law tells us that the two gases will have the same volume if they have the same number of moles at a certain pressure and temperature condition.

At STP, 1 mole of any gas occupies a volume of 22.4 L.

Thus, the molar volume conversion factor can be given as follows:

$$\begin{gathered} 1 \mathrm{mole} \mathrm{of} \mathrm{gas}=22.4 \mathrm{L} \left(\mathrm{STP}\right) \\ \frac{1 \mathrm{mole} \mathrm{gas}}{22.4 \mathrm{L}\left(\mathrm{STP}\right)} \mathrm{and} \frac{22.4 \mathrm{L}\left(\mathrm{STP}\right)}{1 \mathrm{mole} \mathrm{gas}} \end{gathered}$$

Therefore, for 4.00 L CO₂ gas, the number of moles will be given as follows:

$4.00 \cancel{\mathrm{L} {\mathrm{CO}}_2}\times \frac{1 \mathrm{mole} {\mathrm{CO}}_2}{22.4\cancel{ \mathrm{L} {\mathrm{CO}}_{2 }\left(\mathrm{STP}\right)}}=\frac{4.00}{22.4}=0.178 \mathrm{mole}$

Hence, the number of moles of CO₂ gas is 0.178 mole.

Step 1: Part (b) Given Information

The moles of He gas are 0.420 moles

Step 2: Part (b) Explanation

Avogadro law tells us that the two gases will have the same volume if they have the same number of moles at a certain pressure and temperature condition.

At STP, 1 mole of any gas occupies a volume of 22.4 L.

Thus, the molar volume conversion factor can be given as follows:

$$\begin{gathered} 1 \mathrm{mole} \mathrm{of} \mathrm{gas}=22.4 \mathrm{L} \left(\mathrm{STP}\right) \\ \frac{1 \mathrm{mole} \mathrm{gas}}{22.4 \mathrm{L}\left(\mathrm{STP}\right)} \mathrm{and} \frac{22.4 \mathrm{L}\left(\mathrm{STP}\right)}{1 \mathrm{mole} \mathrm{gas}} \end{gathered}$$

Therefore, for 0.420 moles of He gas, the volume will be as follows:

$0.420 \cancel{\mathrm{moles} \mathrm{He}}\times \frac{22.4 \mathrm{L} \mathrm{He}\left(\mathrm{STP}\right)}{1 \cancel{\mathrm{mole} \mathrm{He}}}=9.408 \mathrm{L}$

Hence, the volume of He gas will be 9.408 L.

Step 1: Part (c) Given Information

The mass of O₂ gas is 6.40 g.

Step 2: Part (c) Explanation

Avogadro law tells us that the two gases will have the same volume if they have the same number of moles at a certain pressure and temperature condition.

At STP, 1 mole of any gas occupies a volume of 22.4 L.

Thus, the molar volume conversion factor can be given as follows:

$$\begin{gathered} 1 \mathrm{mole} \mathrm{of} \mathrm{gas}=22.4 \mathrm{L} \left(\mathrm{STP}\right) \\ \frac{1 \mathrm{mole} \mathrm{gas}}{22.4 \mathrm{L}\left(\mathrm{STP}\right)} \mathrm{and} \frac{22.4 \mathrm{L}\left(\mathrm{STP}\right)}{1 \mathrm{mole} \mathrm{gas}} \end{gathered}$$

For oxygen gas, the conversion factor for moles to grams is given as follows:

$$\begin{gathered} 1 \mathrm{mole} \mathrm{of} {\mathrm{O}}_2=32.00 \mathrm{g} \mathrm{of} {\mathrm{O}}_2 \\ \frac{1 \mathrm{mole} {\mathrm{O}}_2}{32.00 \mathrm{g} {\mathrm{O}}_2}\mathrm{and}\frac{32.00 \mathrm{g} {\mathrm{O}}_2}{1 \mathrm{mole} {\mathrm{O}}_2} \end{gathered}$$

Therefore, for 6.40 g of O₂ gas, the volume will be:

$6.40\cancel{ \mathrm{g} {\mathrm{O}}_2}\times \frac{1 \cancel{\mathrm{mole} {\mathrm{O}}_2}}{32.00 \cancel{\mathrm{g} {\mathrm{O}}_2}}\times \frac{22.4 \mathrm{L} {\mathrm{O}}_2\left(\mathrm{STP}\right)}{1 \cancel{\mathrm{mole} {\mathrm{O}}_2}}=\frac{6.40 \times 22.4}{32.00}\mathrm{L}=4.48 \mathrm{L}$

Hence, the volume of O₂ gas will be 4.48 L.

Step 1: Part (d) Given Information

The volume of Ne gas is 11.2 L.

Step 2: Part (d) Explanation

Avogadro law tells us that the two gases will have the same volume if they have the same number of moles at a certain pressure and temperature condition.

At STP, 1 mole of any gas occupies a volume of 22.4 L.

Thus, the molar volume conversion factor can be given as follows:

$$\begin{gathered} 1 \mathrm{mole} \mathrm{of} \mathrm{gas}=22.4 \mathrm{L} \left(\mathrm{STP}\right) \\ \frac{1 \mathrm{mole} \mathrm{gas}}{22.4 \mathrm{L}\left(\mathrm{STP}\right)} \mathrm{and} \frac{22.4 \mathrm{L}\left(\mathrm{STP}\right)}{1 \mathrm{mole} \mathrm{gas}} \end{gathered}$$

Therefore, for 11.2 L of Ne gas, the number of moles can be given as:

$11.2 \cancel{\mathrm{L} \mathrm{Ne}} \times \frac{1 \mathrm{mole} \mathrm{Ne}}{22.4 \cancel{\mathrm{L} \mathrm{Ne} \left(\mathrm{STP}\right)}}=0.5 \mathrm{mole}$

Now, for the calculation of 0.5 moles into grams, the number of moles formula is used.

$$\begin{gathered} \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}} \\ \mathrm{m}=\mathrm{n}\times \mathrm{M} \\ \mathrm{m}=0.5\times 20.1 \\ \mathrm{m}=10.05 \mathrm{g} \end{gathered}$$

Hence, the grams of Ne gas is 10.05 grams.