Question
What is the mass of hydrogen gas produced when 7.43g of zinc is dissolved in 500mL of 1.2M HCl?
Answers
GPT-4o mini
To find the mass of hydrogen gas produced when zinc dissolves in hydrochloric acid (HCl), we'll start with the balanced chemical reaction:
\[
\text{Zn} + 2 \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2
\]
From this equation, we can see that 1 mole of zinc reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.
**Step 1: Calculate moles of zinc (Zn)**
The molar mass of zinc is approximately 65.38 g/mol. To find the number of moles of zinc in 7.43 g:
\[
\text{moles of Zn} = \frac{\text{mass of Zn}}{\text{molar mass of Zn}} = \frac{7.43 \text{ g}}{65.38 \text{ g/mol}} \approx 0.113 \text{ moles of Zn}
\]
**Step 2: Determine the amount of HCl available**
To calculate the moles of HCl present in 500 mL of a 1.2 M solution:
\[
\text{moles of HCl} = \text{Molarity} \times \text{Volume (L)} = 1.2 \text{ mol/L} \times 0.5 \text{ L} = 0.6 \text{ moles of HCl}
\]
**Step 3: Determine the limiting reactant**
The stoichiometry of the reaction dictates that 1 mole of zinc reacts with 2 moles of HCl. Therefore, the moles of HCl required to react with 0.113 moles of zinc is:
\[
\text{moles of HCl required} = 0.113 \text{ moles of Zn} \times 2 = 0.226 \text{ moles of HCl}
\]
Since we have 0.6 moles of HCl available, which is more than enough to react with 0.113 moles of zinc, zinc is the limiting reactant in this reaction.
**Step 4: Calculate moles of hydrogen gas produced**
From the balanced reaction, 1 mole of zinc produces 1 mole of hydrogen gas. Thus, the moles of hydrogen gas produced will equal the moles of zinc reacted:
\[
\text{moles of H}_2 = \text{moles of Zn} = 0.113 \text{ moles}
\]
**Step 5: Calculate mass of hydrogen gas produced**
The molar mass of hydrogen gas (H₂) is approximately 2 g/mol:
\[
\text{mass of H}_2 = \text{moles of H}_2 \times \text{molar mass of H}_2 = 0.113 \text{ moles} \times 2 \text{ g/mol} \approx 0.226 \text{ g}
\]
Thus, the mass of hydrogen gas produced when 7.43 g of zinc is dissolved in 500 mL of 1.2 M HCl is approximately **0.226 g**.
\[
\text{Zn} + 2 \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2
\]
From this equation, we can see that 1 mole of zinc reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.
**Step 1: Calculate moles of zinc (Zn)**
The molar mass of zinc is approximately 65.38 g/mol. To find the number of moles of zinc in 7.43 g:
\[
\text{moles of Zn} = \frac{\text{mass of Zn}}{\text{molar mass of Zn}} = \frac{7.43 \text{ g}}{65.38 \text{ g/mol}} \approx 0.113 \text{ moles of Zn}
\]
**Step 2: Determine the amount of HCl available**
To calculate the moles of HCl present in 500 mL of a 1.2 M solution:
\[
\text{moles of HCl} = \text{Molarity} \times \text{Volume (L)} = 1.2 \text{ mol/L} \times 0.5 \text{ L} = 0.6 \text{ moles of HCl}
\]
**Step 3: Determine the limiting reactant**
The stoichiometry of the reaction dictates that 1 mole of zinc reacts with 2 moles of HCl. Therefore, the moles of HCl required to react with 0.113 moles of zinc is:
\[
\text{moles of HCl required} = 0.113 \text{ moles of Zn} \times 2 = 0.226 \text{ moles of HCl}
\]
Since we have 0.6 moles of HCl available, which is more than enough to react with 0.113 moles of zinc, zinc is the limiting reactant in this reaction.
**Step 4: Calculate moles of hydrogen gas produced**
From the balanced reaction, 1 mole of zinc produces 1 mole of hydrogen gas. Thus, the moles of hydrogen gas produced will equal the moles of zinc reacted:
\[
\text{moles of H}_2 = \text{moles of Zn} = 0.113 \text{ moles}
\]
**Step 5: Calculate mass of hydrogen gas produced**
The molar mass of hydrogen gas (H₂) is approximately 2 g/mol:
\[
\text{mass of H}_2 = \text{moles of H}_2 \times \text{molar mass of H}_2 = 0.113 \text{ moles} \times 2 \text{ g/mol} \approx 0.226 \text{ g}
\]
Thus, the mass of hydrogen gas produced when 7.43 g of zinc is dissolved in 500 mL of 1.2 M HCl is approximately **0.226 g**.