When 200. mL of 0.40 M hydrochloric acid solution is mixed with 3.76 g of aluminum metal, how many moles of hydrogen gas would be produced?

Question

When 200. mL of 0.40 M hydrochloric acid solution is mixed with 3.76 g of aluminum metal, how many moles of hydrogen gas would be produced?

Answer 1

There's a lot going on in this question! We need to understand how the hydrogen from the hydrochloric acid solution converts into hydrogen gas by using a balanced reaction. Then, we can use stoichiometry to determine the limiting reagent. Finally, we'll calculate the amount of H₂ gas produced.

1. The unbalanced equation looks like this: Al _(s) + HCl _(aq) → AlCl3 _(aq) + H2 _(g) ↑
2. The up arrow next to H₂ indicates that they hydrogen gas escapes the solution as gas.
3. To balance the equation, begin with Cl:
4. Al _(s) + 3HCl _(aq) → AlCl3 _(aq) + H2 _(g) ↑
5. 2Al _(s) + 6HCl _(aq) → 2AlCl3 _(aq) + 3H2 _(g) ↑
6. We'll use the stoichiometric relationships given by the balanced equation to find the limiting reagent.
7. 200 mL of 0.40M HCl solution:
8. 200 mL * 1 L/1000 mL * 0.4 mol/1L = 0.08 mol HCl
9. 3.76 g Al:
10. 3.76 g Al * 1 mol/26.98 g Al = 0.139 mol Al
11. Lastly, we determine the limiting reagent. For every 6 mol of HCl, we require 2 mol Al. This means HCl is the limiting reagent.
12. Finally, use the moles of limiting reagent to find the moles of hydrogen gas:
13. 0.08 mol HCl * 3 mol H₂/6 mol HCl = 0.04 mol H₂

Hope that helps!