Respuesta :
Answer : The volume of [tex]H_2O[/tex] 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 = [tex]\frac{\text{ Given mass of Al}}{\text{ Molar mass of Al}}=\frac{6g}{27g/mole}=0.22moles[/tex]
The given balanced reaction is,
[tex]2NaOH+2Al+6H_2O\rightarrow 2NaAl(OH)_4+3H_2[/tex]
From the reaction, we conclude that
2 moles of Al react with the 6 moles of [tex]H_2O[/tex]
0.22 moles of Al react with [tex]\frac{6}{2}\times 0.22=0.66moles[/tex] of [tex]H_2O[/tex]
At STP, 1 mole contains 22.4 L volume
As, 1 mole of [tex]H_2O[/tex] contains 22.4 L volume of [tex]H_2O[/tex]
0.66 moles of [tex]H_2O[/tex] contains [tex](22.4)\times (0.66)=14.784L[/tex] volume of [tex]H_2O[/tex]
Therefore, the volume of [tex]H_2O[/tex] is 14.784 L.
