Question

Aluminum dissolves in an aqueous solution of NaOH according to the following
reaction: 2NaOH + 2Al + 2H2O ! 2NaAlO2 + 3H2 If 84.1 g of NaOH and 51.0 g
of Al react, which is the limiting reagent? How much of the other reagent
remains? What mass of hydrogen is produced?

Answer 1

Answer on Question #82415 – Chemistry – General Chemistry

Question

Aluminum dissolves in an aqueous solution of $NaOH$ according to the following reaction: $2NaOH + 2Al + 2H_2O \rightarrow 2NaAlO_2 + 3H_2$ If $84.1g$ of $NaOH$ and $51.0g$ of $Al$ react, which is the limiting reagent? How much of the other reagent remains? What mass of hydrogen is produced?

Solution

To answer the question, molar masses of $NaOH$, $Al$ and $H_2$ are required.

$$M_{NaOH} \approx 40 \frac{g}{mol}, \quad M_{Al} \approx 27 \frac{g}{mol}, \quad M_{H_2} \approx 2 \frac{g}{mol}.$$

Then it is possible to calculate the chemical amounts of $NaOH$ and $Al$: $n = \frac{m}{M}$.

$$n_{NaOH} = \frac{m_{NaOH}}{M_{NaOH}} = \frac{84.1g}{40 \frac{g}{mol}} = 2.1025mol; \quad n_{Al} = \frac{m_{Al}}{M_{Al}} = \frac{51.0g}{27 \frac{g}{mol}} \approx 1.8889mol.$$

From equation of the reaction it follows, that for every two moles of $NaOH$ it should be two moles of $Al$. However, the chemical amount of $Al$ is less, than the chemical amount of $NaOH$. Therefore, $Al$ is the limiting reagent. After reaction it remains $n_{NaOH(2)} = n_{NaOH} - n_{Al} = 2.1025mol - 1.8889mol = 0.2136mol$ of $NaOH$. It weighs $m_{NaOH} = M_{NaOH} \times n_{NaOH(2)} = 40\frac{g}{mol} \times 0.2136mol = 8.544g$.

For every two moles of $Al$, three moles of $H_2$ are produced. Then, chemical amount of produced $H_2$ equals $\frac{3}{2}$ of the chemical amount of $Al$: $n_{H_2} = \frac{3}{2} \times n_{Al} = \frac{3}{2} \times 1.8889mol = 2.83335mol$.

$$m_{H_2} = M_{H_2} \times n_{H_2} = 2 \frac{g}{mol} \times 2.8335mol = 5.6667g.$$

Answer: $Al$ is the limiting reagent. After reaction it remains $0.2136mol$ or $8.544g$ of $NaOH$. $5.6667g$ of $H_2$ is produced.

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