Question

Given the following two quantities: 0.50 mol of CH4 and 1.0 mol of HCl,

Answer 1

Answer:

(a) HCl

(b) HCl

(c) HCl

(d) HCl

Explanation:

Given: 0.50 mol of CH₄ and 1.0 mol of HCl

Using stoichiometry we can calculate the answers to parts a, b, c, and d.

Part (a)

# of moles × Avogadro's number = # of atoms or molecules

Avogadro's number: 6.02 * 10²³

• $\displaystyle 0.50\ \text{mol CH}_4 \cdot \frac{6.02\cdot 10^2^3 \ \text{atoms CH}_4}{1 \ \text{mol CH}_4} = 3.01 \cdot 10^2^3 \ \text{atoms CH}_4$
• $\displaystyle 1.0\ \text{mol HCl} \cdot \frac{6.02\cdot 10^2^3 \ \text{atoms HCl}}{1 \ \text{mol HCl}} = 6.02 \cdot 10^2^3 \ \text{atoms HCl}$

HCl has more atoms than CH₄.

Part (b)

This is calculated the same way as Part (a); HCl has more molecules than CH₄.

Part (c)

Molar mass of CH₄ = 16.04 g/mol

Molar mass of HCl = 36.458 g/mol

• $\displaystyle 0.50\ \text{mol CH}_4 \cdot \frac{16.04 \ \text{g CH}_4}{1 \ \text{mol CH}_4} = 8.02 \ \text{g CH}_4$
• $\displaystyle 1.0\ \text{mol HCl} \cdot \frac{36.458 \ \text{g HCl}}{1 \ \text{mol HCl}} = 36.458 \ \text{g HCl}$

HCl has a greater mass than CH₄.

Part (d)

Assuming STP:

Molar volume of any gas at STP is 22.4 L/mol.

• $\displaystyle 0.50\ \text{mol CH}_4 \cdot \frac{22.4 \ \text{L CH}_4}{1 \ \text{mol CH}_4} = 11.2 \ \text{L CH}_4$
• $\displaystyle 1.0\ \text{mol HCl} \cdot \frac{22.4 \ \text{L HCl}}{1 \ \text{mol HCl}} = 22.4 \ \text{L HCl}$

HCl has a greater volume than CH₄.