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

First, we need to calculate the number of moles of zinc used in the reaction:

$\text{Molar mass of Zn} = 65.38 \, \text{g/mol}$

Moles of zinc = ( \frac{10.0 , \text{g}}{65.38 , \text{g/mol}} \approx 0.153 , \text{mol} )

According to the balanced equation, 1 mole of zinc produces 1 mole of hydrogen gas. Thus, the moles of hydrogen produced will be the same as the moles of zinc:

• Moles of ( \text{H}_2 ) produced = 0.153 mol

Using the ideal gas law to find the volume at STP (25°C and 100 kPa):

$PV = nRT$

Where:

• ( P = 100 , \text{kPa} )
• ( V ) = volume
• ( n = 0.153 , \text{mol} )
• ( R = 8.314 , \text{J/(mol \cdot K)} ) (or ( 8.314 , \text{kPa} \cdot \text{L/(mol \cdot K)} ))
• ( T = 25 + 273.15 = 298.15 , \text{K} )

Rearranging for V:

$V = \frac{nRT}{P} = \frac{0.153 \, \text{mol} \times 8.314 \, \text{kPa} \cdot \text{L/(mol \cdot K)} \times 298.15 \, \text{K}}{100 \, \text{kPa}} \approx 3.79 \, \text{L}$