Question

How many grams of h2 gas can be produced by the reaction of 63.0 grams of al(s) with an excess of dilute hydrochloric acid in the reaction shown below? 2 al(s) + 6 hcl(aq) → 2 alcl3(aq) + 3 h2(g)?

Answer 1

How many grams of h2 gas can be produced by the reaction of 63.0 grams of al(s) with an excess of dilute hydrochloric acid in the reaction shown below? 2 al(s) + 6 hcl(aq) → 2 alcl3(aq) + 3 h2(g)?

Answer 2

Answer: The mass of hydrogen gas produced by the reaction is 6.9 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

Given mass of aluminium = 63 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{63g}{27g/mol}=2.33mol$

For the given chemical reaction:

$2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)$

Hydrochloric acid is present in excess. So, it is considered as an excess reagent. And, aluminium metal is a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of aluminium metal produces 3 moles of hydrogen gas.

So, 2.33 moles of aluminium metal will produce = $\frac{3}{2}\times 2.33=3.45mol$ of hydrogen gas

Now, calculating the mass of hydrogen gas by using equation 1:

Moles of hydrogen gas = 3.45 moles

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$3.45mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(3.45mol\times 2g/mol)=6.9g$

Hence, the mass of hydrogen gas produced by the reaction is 6.9 grams