Question

When Zn reacts with HCl solution, the products are H₂ gas and ZnCl₂. A volume of 426 mL of H₂ gas is collected over water at a total pressure of 751 mmHg and 16∘C. The vapor pressure of water at 16∘C is 14 mmHg.

Zn(s)+HCl(aq) → H₂(g)+ZnCl₂(aq).
1. What was the partial pressure of the H₂ gas?
2. How many moles of H₂ gas were produced in the reaction?

Answer 1

The partial pressure of H₂ gas is 737 mmHg. The number of moles of H₂ gas produced is 0.0171 moles.

Step-by-step explanation:

In the reaction Zn(s) + HCl(aq) → H₂(g) + ZnCl₂(aq), the partial pressure of the H₂ gas can be calculated by subtracting the vapor pressure of water at 16°C (14 mmHg) from the total pressure of the gas collected (751 mmHg). So, the partial pressure of H₂ gas is 737 mmHg.

To calculate the number of moles of H₂ gas produced, we can use the ideal gas law equation: PV = nRT. We need to calculate the volume of H₂ gas at STP (Standard Temperature and Pressure) and convert it to moles using the molar volume of a gas. The molar volume of a gas at STP is 22.4 L/mol. The volume of H₂ gas collected over water (426 mL) can be converted to liters (0.426 L).

Using PV = nRT, we can rearrange the equation to solve for moles: n = PV/RT.

Plugging in the values, we get n = (737 mmHg - 14 mmHg)(0.426 L)/(0.0821 L·atm/mol·K)(16 + 273 K).

Therefore, the number of moles of H₂ gas produced in the reaction is 0.0171 moles.