Question

What is the minimum amount of 6.9 M H2SO4 necessary to produce 27.8 g of H2 (g) according to the following reaction? 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)

Answer 1

Answer : The minimum amount of 6.9 M
`H_2SO_4` needed is, 2.0 L

Explanation :

The given chemical reaction is:

`2Al(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2(g)`

First we have to calculate the moles of
`H_2`

`\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}`

Molar mass of
`H_2` = 2 g/mol

`\text{Moles of }H_2=(27.8g)/(2g/mol)=13.9mol`

Now we have to calculate the moles of
`H_2SO_4`

From the balanced chemical reaction we conclude that,

As, 3 moles of
`H_2` produced from 3 moles of
`H_2SO_4`

So, 13.9 moles of
`H_2` produced from 13.9 moles of
`H_2SO_4`

Now we have to calculate the mass of
`H_2SO_4`

`\text{Concentration of }H_2SO_4=\frac{\text{Moles of }H_2SO_4}{\text{Volume of }H_2SO_4}`

`6.9M=\frac{13.9mol}{\text{Volume of }H_2SO_4}`

`\text{Volume of }H_2SO_4=2.0L`

Thus, the minimum amount of 6.9 M
`H_2SO_4` needed is, 2.0 L