Answer : The mass percent of Al in the alloy is 85.9 %.
Explanation : Given,
Mass of sample of an alloy = 1.1181 g
Mass of [tex]H_2[/tex] = 0.1068 g
Molar mass of [tex]H_2[/tex] = 2 g/mol
Molar mass of [tex]Al[/tex] = 27 g/mol
First we have to calculate the moles of [tex]H_2[/tex].
[tex]\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}[/tex]
[tex]\text{Moles of }H_2=\frac{0.1068g}{2g/mol}=0.0534mol[/tex]
Now we have to calculate the moles of [tex]Al[/tex]
The balanced chemical equation is:
[tex]2Al+2NaOH+6H_2O\rightarrow 2Na[Al(OH)_4]+3H_2[/tex]
From the reaction, we conclude that
As, 3 moles of [tex]H_2[/tex] produced from 2 moles of [tex]Al[/tex]
So, 0.0534 mole of [tex]H_2[/tex] produced from [tex]\frac{2}{3}\times 0.0534=0.0356[/tex] mole of [tex]CaCl_2[/tex]
Now we have to calculate the mass of [tex]Al[/tex]
[tex]\text{ Mass of }Al=\text{ Moles of }Al\times \text{ Molar mass of }Al[/tex]
[tex]\text{ Mass of }Al=(0.0356moles)\times (27g/mole)=0.9612g[/tex]
Now we have to calculate the mass percent of Al in the alloy.
Mass percent of Al in alloy = [tex]\frac{\text{Mass of Al}}{\text{Mass of sample of an alloy}}\times 100[/tex]
Mass percent of Al in alloy = [tex]\frac{0.9612g}{1.1181g}\times 100[/tex]
Mass percent of Al in alloy = 85.9%
Therefore, the mass percent of Al in the alloy is 85.9 %.