Question

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

excess of sodium hydroxide solution. In the reaction, only the aluminum reacts with the sodium
hydroxide solution:
2 Al + 2 NaOH + 6 H202 Na[Al(OH)4] + 3 H2
If 0.1068 g of H2 is produced, what is the mass percent of aluminum in the alloy?

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=(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
`(2)/(3)* 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* \text{ Molar mass of }Al`

`\text{ Mass of }Al=(0.0356moles)* (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}}* 100`

Mass percent of Al in alloy =
`(0.9612g)/(1.1181g)* 100`

Mass percent of Al in alloy = 85.9%

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