Question

All faculty members are happy to see students help each other. Dumbledore is particularly pleased with Hermione. Though, it should be mentioned that students should not simply copy off each other. You will not learn anything that way. Snape glares at Ron... Ron slouches in his chair. Snape thinks it’s time for a harder problem. How many milligrams of magnesium reacts with excess HCl to produce 31.2 mL of hydrogen gas at 754 Torr and 25.0◦C. The hydrogen is produced by the following reaction: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) Express your answer in milligrams.

Answer 1

`\large \boxed{\textbf{30.7 mg}}`

Step-by-step 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:

`\text{p} = \text{754 torr} * \frac{\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}*\text{0.0312 L}  & = & n * 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^(-1)\text{mol}^(-1) * \text{ 298.15 K}\\0.03095 & = & 24.47n \text{ mol}^(-1)\\n & = & \frac{0.03095}{24.47 \text{ mol}^(-1)}\\\\& = & 1.263 * 10^(-3) \text{ mol}\\\end{array}`

(b) Calculate the moles of Mg

MM: 24.30

Mg + 2HCl ⟶ MgCl₂ + H₂

n/mol: 1.263 × 10⁻³

`\text{Moles of Mg} = 1.263 * 10^(-3)\text{ mol H}_(2) * \frac{\text{1 mol Mg}}{\text{1 mol H}_(2) } = 1.263 * 10^(-3)\text{ mol Mg}`

(c) Calculate the mass of Mg

`\text{Mass of Mg} = 1.263 * 10^(-3)\text{ mol Mg} * \frac{\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 $\large \boxed{\textbf{30.7 mg}}$}`