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

Reaction: Al + 3HCl → AlCl₃ + (3/2) H₂

1) LR: See if ratio of HCl/Al is greater or less than 3

Moles of Al = 25 g * 1mol/26.98 g = .9266

Moles of HCl = 119 g * 1 mole/36.46 g = 3.364 (clearly an excess of HCl or LR is Al)

2) .9266 moles Al * (3/2) H₂/1 Al = 1.390 moles H₂

Now use IGL to find V = nRT/P with R = .0821, T=305 K and P = 1 atm

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