Question

A 3.50 g sample of aluminum reacts with excess H2SO4. What volume of H2 gas is produced at a pressure of 735 mmHg and a temperature of 22C?

Answer 1

Solution- Given mass of Al il 3.50g Atomic mass of Al = 27 u = Moles of Al Given mass mo Atomic mass = 3. so 27 = 0.130 mal # The reaction taking place is 2 Al + 3 H2SO4 Al [sou), + 3 H2 73 mol H2 2 mol Al react to produce 0.130 mol Al produced .: 3 mol H₂ x 0.130 2 0.1.95 moles H₂ gas # 735 atm -0.967 atm = 298 K Pressure (P) = 735 mmHg 760 Temperature (T) = 22°C +273 moles of Hg = 0.195 mol Ra 0.0821 L-atm. mol-k Putting values in ideal gas equation CPV = nRT) we get v nRT 0.195 X 0.0821 x 295 P 0.967 v 4.7228 4.884 L 0.967 4.90 L

For calculating volume of H2 gas ideal gas equation is used after finding moles of H2 produced.

The volume of H2 gas = 4.884 L

~ 4.90 L