A gas is produced when 10.0 g of zinc is placed in 0.50 L of 0.20 mol L$^{-1}$ nitric acid - HSC - SSCE Chemistry - Question 26 - 2010 - Paper 1

To determine the number of moles of zinc present, we can use the formula:

$ext{Moles} = \frac{ ext{mass}}{ ext{molar mass}}$

The molar mass of zinc (Zn) is approximately 65.38 g/mol. Thus, the moles of zinc is:

$$ext{Moles of Zn} = \frac{10.0 ext{ g}}{65.38 ext{ g/mol}} \approx 0.153 ext{ mol}$$

According to the balanced chemical equation, 1 mole of zinc produces 1 mole of hydrogen gas. Therefore, the moles of hydrogen produced will also be approximately 0.153 mol.

Now, we can use the Ideal Gas Law to calculate the volume of gas produced at 25°C (298 K) and 100 kPa (0.1 MPa):

$$PV = nRT$$

Where:

• $P = 100 ext{ kPa} = 0.1 ext{ MPa}$
• $n = 0.153 ext{ mol}$
• $R = 8.314 ext{ J/(mol K)}$ or $0.08314 ext{ L kPa/(mol K)}$
• $T = 298 ext{ K}$

Rearranging the equation to solve for volume (V):

$$v = \frac{nRT}{P}$$

Calculating the volume:

$$v = \frac{0.153 ext{ mol} \times 0.08314 ext{ L kPa/(mol K)} \times 298 ext{ K}}{0.1 ext{ MPa}} \approx 3.79 ext{ L}$$

Thus, the volume of gas produced is approximately 3.79 L.