Question

2Al + 6HCl → 2AlCl3 + 3H2 If the chemical reaction produces 129 grams of AlCl3, how many grams of H2 are also produced?

Answer 1

So,

Our conceptual plan is as follows:

g AlCl3 --> mol AlCl3 --> mol H2 --> g H2

$\frac{129g\ AlCl_3}{133.4g/mol}\ *\frac{3mol\ H_2}{2mol\ AlCl_3}\ *2.016g/mol \\ \\=2.92 g\ H_2$

Hope this helps!

Answer 2

Answer:

B. 2.92

Explanation:

• It is a stichiometry problem.

• From the balanced given reaction: 2Al + 6HCl → 2AlCl₃ + 3H₂,

• It is clear that 2.o moles of Al reacts with 6.0 moles of HCl to produce 2.0 moles of AlCl₃ and 3.0 moles of H₂.

• Herein, we are concerned with the two products that the reaction results 2.0 moles of AlCl₃ with 2.0 moles of H₂.

• So, we should convert the amount of grams of AlCl₃ (129.0 g) to number of moles (n) using the relation:

n = mass / molar mass,

∴ n of AlCl₃ = (129.0 g) / (133.34 g/mol) = 0.967 mol.

Using cross multiplication:

2.0 moles of AlCl₃ produced with → 3.0 moles of H₂, from the stichiometry.

0.967 moles of AlCl₃ produced with → ??? moles of H₂.

∴ the number of moles of H₂ = (0.967)(3.0) / (2.0) = 1.45 mol.

• Now, we can get the grams of H₂:

∴ The grams of H₂ = n x molar mass = (1.45 mol)(2.01588 g/mol) = 2.92 g.