Question

Sodium hydroxide reacts with aluminum and water to produce hydrogen gas:2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g)What mass of hydrogen gas would be formed from a reaction of 1.33 g Al and 4.25 g NaOH in water?

Answer 1

Answer: 0.147 grams

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} Al=\frac{1.33g}{27g/mol}=0.049moles$

$\text{Moles of} NaOH=\frac{4.25g}{40g/mol}=0.106moles$

$2Al(s)+2NaOH(aq)+6H_2O(l)\rightarrow Na(Al(OH)_4(aq)+3H_2(g)$

According to stoichiometry :

2 moles of $Al$ require = 2 moles of $NaOH/tex]Thus 0.049 moles of [tex]Al$ will require=$\frac{2}{2}\times 0.049=0.049moles$  of $NaOH$

Thus $Al$ is the limiting reagent as it limits the formation of product and $NaOH$ is the excess reagent.

As 2 moles of $Al$ give = 3 moles of $H_2$

Thus 0.049 moles of $Al$ give =$\frac{3}{2}\times 0.049=0.0735moles$  of $H_2$

Mass of $H_2=moles\times {\text {Molar mass}}=0.0735moles\times 2g/mol=0.147g$

Thus 0.147 g of $H_2$ will be produced from the given masses of both reactants.