Question

The action of some commercial drain cleaners is based on the following reaction: 2 NaOH(s) + 2 Al(s) + 6 H2O(l) → 2 NaAl(OH)4(s) + 3 H2(g) What is the volume of H2 gas formed at STP when 6.32 g of Al reacts with excess NaOH?

Answer 1

Answer : The volume of $H_2$ gas formed at STP is 7.86 liters.

Explanation :

The balanced chemical reaction will be:

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

First we have to calculate the moles of $Al$.

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

Molar mass of Al = 27 g/mole

$\text{Moles of }Al=\frac{6.32g}{27g/mole}=0.234mole$

Now we have to calculate the moles of $H_2$ gas.

From the reaction we conclude that,

As, 2 mole of $Al$ react to give 3 mole of $H_2$

So, 0.234 moles of $Al$ react to give $\frac{0.234}{2}\times 3=0.351$ moles of $H_2$

Now we have to calculate the volume of $H_2$ gas formed at STP.

As, 1 mole of $H_2$ gas contains 22.4 L volume of $H_2$  gas

So, 0.351 mole of $H_2$ gas contains $0.351\times 22.4=7.86L$ volume of $H_2$ gas

Therefore, the volume of $H_2$ gas formed at STP is 7.86 liters.