Question

If you have 100 g of NaOH and 100 g of Al to perform the reaction, how many grams of H2 will you produce?
6NaOH(aq) + 2Al(s) ® 2NA3AlO3(s) + 3H2(g)

Answer 1

Answer on Question #51553 - Chemistry – Other

Question

If you have 100 g of NaOH and 100 g of Al to perform the reaction, how many grams of H₂ will you produce?

$$6 \mathrm{NaOH}(aq) + 2 \mathrm{Al}(s) = 2 \mathrm{NA_3AlO_3}(s) + 3 \mathrm{H_2}(g)$$

Answer:

Molar masses of the reactants equal:

$$M(\mathrm{NaOH}) = 40 \ \mathrm{g/mol}, \quad M(\mathrm{Al}) = 27 \ \mathrm{g/mol}$$

Number of moles of the reactants are:

$$n(Al) = \frac{m(Al)}{M(Al)} = \frac{100}{27} = 3.7 \ \mathrm{mol}$$ $$n(\mathrm{NaOH}) = \frac{m(\mathrm{NaOH})}{M(\mathrm{NaOH})} = \frac{100}{40} = 2.5 \ \mathrm{mol}$$

Then we make a proportion:

- 2 moles of Al react with 6 moles of NaOH

- 3.7 moles of Al – x moles of NaOH

$$x = \frac{3.7 \cdot 6}{2} = 11.1 \ \mathrm{moles} \ \text{of} \ \mathrm{NaOH} \ \text{should} \ \text{react} \ \text{with} \ 3.7 \ \mathrm{moles} \ \text{of} \ \mathrm{Al}$$

There are only 2.5 moles of sodium hydroxide, therefore it is the limiting reactant.

We need to make another proportion to calculate the mass of H₂ that could be produced by the chemical reaction:

- 6 moles of NaOH produce 3 moles of H₂

- 2.5 moles of NaOH – x moles of H₂

$$x = \frac{2.5 \cdot 3}{6} = 1.25 \ \mathrm{moles} \ \text{of} \ \mathrm{H_2} \ \text{could} \ \text{be} \ \text{produced}$$

The mass of H₂ equals:

$$m(H_2) = n(H_2) \cdot M(H_2) = 1.25 \cdot 2 = 2.5 \ g$$

Answer: 2.5 g of H₂

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