Answer: 290 g of aluminium sulphate is produced.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} H_2SO_4=\frac{250g}{98g/mol}=2.55moles[/tex]
The balanced chemical reaction is:
[tex]2Al(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2(g)[/tex]
According to stoichiometry :
3 moles of [tex]H_2SO_4[/tex] produce = 1 mole of [tex]Al_2(SO_4)_3[/tex]
Thus 2.55 moles of [tex]H_2SO_4[/tex] will require=[tex]\frac{1}{3}\times 2.55=0.85moles[/tex] of [tex]Al_2(SO_4)_3[/tex]
Mass of [tex]Al_2(SO_4)_3=moles\times {\text {Molar mass}}=0.85moles\times 342g/mol=290g[/tex]
Thus 290 g of aluminium sulphate is produced.
