Question

5. A 1.1181-g sample of an alloy (a mixture) of aluminum and magnesium was treated with an

Answer 1

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 $H_2$ = 0.1068 g

Molar mass of $H_2$ = 2 g/mol

Molar mass of $Al$ = 27 g/mol

First we have to calculate the moles of $H_2$.

$\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}$

$\text{Moles of }H_2=\frac{0.1068g}{2g/mol}=0.0534mol$

Now we have to calculate the moles of $Al$

The balanced chemical equation is:

$2Al+2NaOH+6H_2O\rightarrow 2Na[Al(OH)_4]+3H_2$

From the reaction, we conclude that

As, 3 moles of $H_2$ produced from 2 moles of $Al$

So, 0.0534 mole of $H_2$ produced from $\frac{2}{3}\times 0.0534=0.0356$ mole of $CaCl_2$

Now we have to calculate the mass of $Al$

$\text{ Mass of }Al=\text{ Moles of }Al\times \text{ Molar mass of }Al$

$\text{ Mass of }Al=(0.0356moles)\times (27g/mole)=0.9612g$

Now we have to calculate the mass percent of Al in the alloy.

Mass percent of Al in alloy = $\frac{\text{Mass of Al}}{\text{Mass of sample of an alloy}}\times 100$

Mass percent of Al in alloy = $\frac{0.9612g}{1.1181g}\times 100$

Mass percent of Al in alloy = 85.9%

Therefore, the mass percent of Al in the alloy is 85.9 %.