Q.8.51 - 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 O₂ in 44.8 L of O₂  gas.

b. the volume in liters, occupied by 2.50 moles of N₂ gas.

c. the volume, in liters, occupied by 50.0 g of Ar gas.

d. the number of grams of H₂ in 16.20 mL of H₂ gas.

Step-by-Step Solution

Answer

a. The number of moles is 2 moles.

b. The volume is 56 L.

c. The volume is  28 L.

d. The number of grams is 0.00072 grams.

Step 1: Part (a) Given Information

The volume of O₂ gas is 44.8 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:

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

therefore, for 44.8 L O₂ gas, the number of moles will be given as follows:

$44.8\cancel{ L O_2}\times \frac{1 \mathrm{mole} \mathrm{gas}}{22.4\cancel{ \mathrm{L} O_2 (\mathrm{STP}})}=\frac{44.8 }{22.4}\mathrm{mole} \mathrm{gas} =2 \mathrm{mole} \mathrm{gas}$

Hence, the number of moles of O₂ gas are 2 moles.

Step 1: Part (b) Given Information

The mole of  N₂ gas is 2.50 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:

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

Therefore, for 2.50 moles of N₂ gas, the volume will be as follows:

data-custom-editor="chemistry" $2.50\cancel{ \mathrm{mole} {\mathrm{N}}_2}\times \frac{22.4 \mathrm{L} {\mathrm{N}}_2 \left(\mathrm{STP}\right)}{1 \cancel{\mathrm{mole} {\mathrm{N}}_2}}=56 \mathrm{L}$

Hence, the volume in liters will be 56 L.

Step 1: Part (c) Given Information

The mass of Ar gas is 50.0 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:

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

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

$$\begin{gathered} 1 \mathrm{mole} \mathrm{of} \mathrm{Ar}=40.0 \mathrm{g} \mathrm{of} \mathrm{Ar} \\ \frac{40.0 \mathrm{g} \mathrm{Ar}}{1 \mathrm{mole} \mathrm{Ar}}=\frac{1 \mathrm{mole} \mathrm{Ar}}{40.0 \mathrm{g} \mathrm{Ar}} \end{gathered}$$

Thus, for 50.0 g of Ar, the volume will be given as follows:

$50.0 \cancel{\mathrm{g} \mathrm{Ar}}\times \frac{1 \cancel{\mathrm{mole} \mathrm{Ar}}}{40.0\cancel{ \mathrm{g} \mathrm{Ar}}}\times \frac{22.4 \mathrm{L} \mathrm{Ar} \left(\mathrm{STP}\right)}{1 \cancel{\mathrm{mole} \mathrm{Ar}}}=\frac{22.4\times 50.0}{40.0}\mathrm{L}=28 \mathrm{L}$

Hence, the volume in liters will be 28 L.

Part (d) Step 1: Given Information

The volume of H₂ gas is 16.20 mL.

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:

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

Therefore, for 16.20 mL of H₂ gas, the number of grams will be:

$16.20 \cancel{\mathrm{mL}}\times \frac{1 \cancel{\mathrm{L}}}{1000 \cancel{\mathrm{mL}}}\times \frac{1 \mathrm{mole} {\mathrm{H}}_2}{22.4 \cancel{\mathrm{L} \left(\mathrm{STP}\right)}}=0.00072 \mathrm{mole}$

Now, for the calculation of Grams of H₂ gas, using the number of moles relation:

$$\begin{gathered} \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}} \\ \mathrm{m}=\mathrm{n}\times \mathrm{M} \\ \mathrm{m}=0.00072\times 1 \\ \mathrm{m}=0.0072 \mathrm{g} \end{gathered}$$

Hence, the grams of H₂ gas is 0.00072 grams.