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=(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
`(0.234)/(2)* 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* 22.4=7.86L` volume of
`H_2` gas

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