Question

What volume of H2O is formed at stp when 6.0g of Al is treated with excess NaOH?

Answer 1

Answer : The volume of $H_2O$ is 14.784 L.

Solution : Given,

Mass of Aluminium = 6 g

Molar mass of Aluminium = 27 g/mole

First we have to calculate the moles of aluminium.

Moles of Al = $\frac{\text{ Given mass of Al}}{\text{ Molar mass of Al}}=\frac{6g}{27g/mole}=0.22moles$

The given balanced reaction is,

$2NaOH+2Al+6H_2O\rightarrow 2NaAl(OH)_4+3H_2$

From the reaction, we conclude that

2 moles of Al react with the 6 moles of $H_2O$

0.22 moles of Al react with $\frac{6}{2}\times 0.22=0.66moles$ of $H_2O$

At STP,  1 mole contains 22.4 L volume

As, 1 mole of $H_2O$ contains 22.4 L volume of $H_2O$

0.66 moles of $H_2O$ contains $(22.4)\times (0.66)=14.784L$ volume of $H_2O$

Therefore, the volume of $H_2O$ is 14.784 L.