My question concerns the stoichiometry of gases.

Question

Solid aluminum reacts with an aqueous solution of HCl to produce an aqueous solution of AlCl3 and hydrogen gas. If 110.0 g HCl are in the solution and 25.00 g Al are added, determine the following:

a. the limiting reactant

b. the volume of H2, produced at atmospheric pressure and 32 degrees C.

Answer 1

2Al(s) + 6HCl(aq) ==> 2AlCl₃(aq) + 3H₂(g) .. balanced equation

a). Limiting reactant. One easy way to find limiting reactant is to divide moles of each reactant by the corresponding coefficient in the balanced equation. Which ever value is less represents the limiting reactant.

For Al: 25.00 g Al x 1 mol Al / 26.98 g = 0.9266 mol Al (÷2->0.463)

For HCl: 110.0 g HCl x 1 mol / 36.46 g = 3.017 mol HCl (÷6->0.503)

Since 0.463 is less than 0.503, Al is the limiting reactant.

b). Volume of H₂ gas produced at atmospheric pressure and 32ºC. Use the moles of Al (limiting reactant) to determine moles of H₂ gas produced. Then use the ideal gas law to find the volume of H₂ gas.

moles H₂ gas produced = 0.9266 mol Al x 3 mol H₂ / 2 mol Al = 1.390 mol H₂ gas

PV = nRT .. Ideal Gas Law

P = pressure = 1 atm

V = volume of H₂ = ?

n = mols H₂ = 1.390 mol

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 32 + 273 = 305K

Solve for V (volume):

V = nRT/P = (1.390)(0.0821)(305) / 1

V = 34.8 L