Answer : The minimum amount of 6.9 M [tex]H_2SO_4[/tex] needed is, 2.0 L
Explanation :
The given chemical reaction is:
[tex]2Al(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2(g)[/tex]
First we have to calculate the moles of [tex]H_2[/tex]
[tex]\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}[/tex]
Molar mass of [tex]H_2[/tex] = 2 g/mol
[tex]\text{Moles of }H_2=\frac{27.8g}{2g/mol}=13.9mol[/tex]
Now we have to calculate the moles of [tex]H_2SO_4[/tex]
From the balanced chemical reaction we conclude that,
As, 3 moles of [tex]H_2[/tex] produced from 3 moles of [tex]H_2SO_4[/tex]
So, 13.9 moles of [tex]H_2[/tex] produced from 13.9 moles of [tex]H_2SO_4[/tex]
Now we have to calculate the mass of [tex]H_2SO_4[/tex]
[tex]\text{Concentration of }H_2SO_4=\frac{\text{Moles of }H_2SO_4}{\text{Volume of }H_2SO_4}[/tex]
[tex]6.9M=\frac{13.9mol}{\text{Volume of }H_2SO_4}[/tex]
[tex]\text{Volume of }H_2SO_4=2.0L[/tex]
Thus, the minimum amount of 6.9 M [tex]H_2SO_4[/tex] needed is, 2.0 L
