The volume occupied by 17g of H2S at s.t.p. is [H = 1.00, S = 32.0, Molar volume = 22.4 dm3]

The volume occupied by 17g of H₂S at s.t.p. is [H = 1.00, S = 32.0, Molar volume = 22.4 dm³]

Answer Details

To solve this problem, we need to use the ideal gas law, which relates the volume, pressure, temperature, and number of moles of a gas. At standard temperature and pressure (STP), which is defined as 0°C and 1 atmosphere of pressure, one mole of any gas occupies 22.4 dm³ of volume. First, we need to calculate the number of moles of H2S present in 17g of the gas. To do this, we divide the mass of the gas by its molar mass. Molar mass of H2S = (2 × atomic mass of H) + atomic mass of S = (2 × 1.00 g/mol) + 32.0 g/mol = 34.0 g/mol Number of moles of H2S = mass of H2S ÷ molar mass of H2S = 17g ÷ 34.0 g/mol = 0.5 mol Now, we can use the ideal gas law to calculate the volume of 0.5 mol of H2S at STP. PV = nRT where P = 1 atm, V is the volume we want to find, n = 0.5 mol, R is the gas constant (0.0821 L atm/K mol), and T = 273 K. Solving for V, we get: V = nRT/P = (0.5 mol) × (0.0821 L atm/K mol) × (273 K) ÷ (1 atm) = 11.2 L or dm³ Therefore, the volume occupied by 17g of H2S at STP is 11.2 dm³. Answer: 11.2 dm³.