Answer:
[tex]\large \boxed{\textbf{30.7 mg}}[/tex]
Explanation:
(a) Calculate the moles of H₂
We can use the ideal gas law:
pV = nRT
Data:
p = 754 torr
V = 31.2 mL
R = 0.082 06 L·atm·K⁻¹mol⁻¹
T = 25.0 °C K
Calculation:
[tex]\text{p} = \text{754 torr} \times \dfrac{\text{1 atm}}{\text{760 torr}} = \text{0.9921 atm}\\\\V = \text{ 31.2 mL} = \text{0.0312 L}\\T = \text{(25.0 + 273.15) K} = \text{298.15 K}\\\\\begin{array}{rcl}\text{0.9921 atm}\times\text{0.0312 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{ 298.15 K}\\0.03095 & = & 24.47n \text{ mol}^{-1}\\n & = & \dfrac{0.03095}{24.47 \text{ mol}^{-1}}\\\\& = & 1.263 \times 10^{-3} \text{ mol}\\\end{array}[/tex]
(b) Calculate the moles of Mg
MM: 24.30
Mg + 2HCl ⟶ MgCl₂ + H₂
n/mol: 1.263 × 10⁻³
[tex]\text{Moles of Mg} = 1.263 \times 10^{-3}\text{ mol H}_{2} \times \dfrac{\text{1 mol Mg}}{\text{1 mol H}_{2} } = 1.263 \times 10^{-3}\text{ mol Mg}[/tex]
(c) Calculate the mass of Mg
[tex]\text{Mass of Mg} = 1.263 \times 10^{-3}\text{ mol Mg} \times \dfrac{\text{24.30 g Mg}}{\text{1 mol Mg}} = \text{0.0307 g Mg } = \textbf{30.7 mg Mg}\\\\\text{The mass of Mg that reacts is $\large \boxed{\textbf{30.7 mg}}$}[/tex]
