Question

Problem 11.111 Review IC Part B Solid aluminum reacts with aqueous H, SO to form H, gas and aluminum sulfate. When a sample of Al is allowed to react, 370 mL of gas is collected over water at 23 °C, at a pressure of 790 mmHg. At 23 °C, the vapor pressure of water is 21 mmHg: 2Al(s) + 3H, SO. (ag) +3H2(g) + Al(SO4)3(aq) How many moles of H, were produced? Express your answer with the appropriate units. Value Units Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining Part How many grams of Al were reacted? Express your answer with the appropriate units. Previous

Answer 1

Part B)

Pressure of H2 = P of wet gas - P of water vapor = 790-21= 769mmHg = 769mm x 1 atm/760 mm= 1.012 atm

V of H2 = 370mL = 370mL x 1L /1000mL = 0.370L

T = 23 C = 23 + 273 = 296 K

Using the ideal gas equation PV = nRT

n , number of moles of H2 = PV/RT = 1.012 atmx 0.370L / 0.0821L.atm/mol.K x 296K

=0.01540 mol

part C

From the equation balanced,

2 mole of Al produces 3 mol of H2

To produce 0.01540mol H2 , moles of Al reuired = 0.0154 mol H2 x 2 mol Al / 3 mol H2

=0.01027 mol Al

Mass of AL needed = mol x molar mass = 0.01027mol x 27g/mol =0.277g