Question

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

NaOH + Al + H2O —-> NaAl(OH)4 + H2 (g)

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}}=(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
`(6)/(2)* 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)* (0.66)=14.784L` volume of
`H_2O`

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