QuestionMay 16, 2025

What volume of H_(2) gas, in mL, is produced from 0.260 g Mg reacting with an excess of HCl at a pressure of 1.28 atm and a temperature of 312K? Mg(s)+2HCl(aq)arrow MgCl_(2)(aq)+H_(2)(g)

What volume of H_(2) gas, in mL, is produced from 0.260 g Mg reacting with an excess of HCl at a pressure of 1.28 atm and a temperature of 312K? Mg(s)+2HCl(aq)arrow MgCl_(2)(aq)+H_(2)(g)
What volume of H_(2) gas, in mL, is produced from 0.260 g Mg reacting with an excess of HCl at a pressure of 1.28 atm and a temperature of 312K? Mg(s)+2HCl(aq)arrow MgCl_(2)(aq)+H_(2)(g)

Solution
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Answer

224.5 mL Explanation 1. Calculate moles of Mg Molar mass of Mg is 24.305 g/mol. Moles of Mg = \frac{0.260 \text{ g}}{24.305 \text{ g/mol}}. 2. Determine moles of H_2 From the balanced equation, 1 mole of Mg produces 1 mole of H_2. Thus, moles of H_2 = moles of Mg. 3. Use Ideal Gas Law to find volume **PV = nRT**. Solve for V: V = \frac{nRT}{P}. Use R = 0.0821 \text{ L atm/mol K}, T = 312 \text{ K}, and P = 1.28 \text{ atm}. 4. Convert volume to mL Multiply volume in liters by 1000 to convert to mL.

Explanation

1. Calculate moles of Mg<br /> Molar mass of Mg is 24.305 g/mol. Moles of Mg = $\frac{0.260 \text{ g}}{24.305 \text{ g/mol}}$.<br /><br />2. Determine moles of $H_2$<br /> From the balanced equation, 1 mole of Mg produces 1 mole of $H_2$. Thus, moles of $H_2$ = moles of Mg.<br /><br />3. Use Ideal Gas Law to find volume<br /> **$PV = nRT$**. Solve for $V$: $V = \frac{nRT}{P}$. Use $R = 0.0821 \text{ L atm/mol K}$, $T = 312 \text{ K}$, and $P = 1.28 \text{ atm}$.<br /><br />4. Convert volume to mL<br /> Multiply volume in liters by 1000 to convert to mL.
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