Zn reacts with hydrochloric acid. Zn(s)+ 2 HCl(aq)→ ZnCl₂(aq)+ H₂(g) What volume of 3.05 M HCl(aq)will react with 25.0 g Zn(s)?

Question

Quizplus · Accepted Answer

First, calculate the moles of Zn: $ \frac{25.0 \, \text{g Zn}}{65.38 \, \text{g/mol}} = 0.382 \, \text{mol Zn} $. According to the balanced equation, 1 mole of Zn reacts with 2 moles of HCl. Therefore, $0.382 \, \text{mol Zn} \times 2 = 0.764 \, \text{mol HCl}$ is needed. To find the volume of 3.05 M HCl required: $ \frac{0.764 \, \text{mol HCl}}{3.05 \, \text{mol/L}} = 0.251 \, \text{L} $.

Zn Reacts with Hydrochloric Acid