QuestionJuly 17, 2025

6) When 2,400 mg zinc is added to hydrochloric acid, 450 mL of hydrogen gas forms at a temperature of 30^circ C What is the pressure of the gas (in atm)? underline ( )Zn(s)+underline ( )HCl(aq)arrow underline ( )ZnCl_(2)(aq)+underline ( )H_(2)(g)

6) When 2,400 mg zinc is added to hydrochloric acid, 450 mL of hydrogen gas forms at a temperature of 30^circ C What is the pressure of the gas (in atm)? underline ( )Zn(s)+underline ( )HCl(aq)arrow underline ( )ZnCl_(2)(aq)+underline ( )H_(2)(g)
6) When 2,400 mg zinc is added to hydrochloric acid, 450 mL of hydrogen gas forms at a temperature of 30^circ C What is the pressure of the gas (in atm)? underline ( )Zn(s)+underline ( )HCl(aq)arrow underline ( )ZnCl_(2)(aq)+underline ( )H_(2)(g)

Solution
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Answer

The pressure of the gas is approximately 0.081 atm. Explanation 1. Convert mass of zinc to moles Molar mass of Zn is 65.38 g/mol. Convert 2400 mg to grams: 2400 \, \text{mg} = 2.4 \, \text{g}. Calculate moles: \text{moles of Zn} = \frac{2.4}{65.38}. 2. Determine moles of hydrogen gas From the balanced equation, 1 mole of Zn produces 1 mole of H_2. Thus, moles of H_2 = moles of Zn. 3. Use Ideal Gas Law to find pressure Convert temperature to Kelvin: T = 30 + 273.15 = 303.15 \, \text{K}. Volume V = 450 \, \text{mL} = 0.450 \, \text{L}. Use **Ideal Gas Law**: PV = nRT, where R = 0.0821 \, \text{L atm/mol K}. Solve for P: P = \frac{nRT}{V}.

Explanation

1. Convert mass of zinc to moles<br /> Molar mass of Zn is 65.38 g/mol. Convert 2400 mg to grams: $2400 \, \text{mg} = 2.4 \, \text{g}$. Calculate moles: $\text{moles of Zn} = \frac{2.4}{65.38}$.<br /><br />2. Determine moles of hydrogen gas<br /> From the balanced equation, 1 mole of Zn produces 1 mole of $H_2$. Thus, moles of $H_2$ = moles of Zn.<br /><br />3. Use Ideal Gas Law to find pressure<br /> Convert temperature to Kelvin: $T = 30 + 273.15 = 303.15 \, \text{K}$. Volume $V = 450 \, \text{mL} = 0.450 \, \text{L}$. Use **Ideal Gas Law**: $PV = nRT$, where $R = 0.0821 \, \text{L atm/mol K}$. Solve for $P$: $P = \frac{nRT}{V}$.
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