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Sagot :
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
- [tex]\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[/tex]
- [tex]\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}[/tex]
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
- [tex]\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[/tex]
- [tex]\displaystyle 1.0\ \text{mol HCl} \cdot \frac{36.458 \ \text{g HCl}}{1 \ \text{mol HCl}} = 36.458 \ \text{g HCl}[/tex]
HCl has a greater mass than CH₄.
Part (d)
Assuming STP:
Molar volume of any gas at STP is 22.4 L/mol.
- [tex]\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[/tex]
- [tex]\displaystyle 1.0\ \text{mol HCl} \cdot \frac{22.4 \ \text{L HCl}}{1 \ \text{mol HCl}} = 22.4 \ \text{L HCl}[/tex]
HCl has a greater volume than CH₄.
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