Question

250 g H 2 SO 4 completely reacted with aluminum?

Answer 1

Answer: 290 g of aluminium sulphate is produced.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} H_2SO_4=\frac{250g}{98g/mol}=2.55moles$

The balanced chemical reaction is:

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

According to stoichiometry :

3 moles of $H_2SO_4$ produce  = 1 mole of $Al_2(SO_4)_3$

Thus 2.55 moles of $H_2SO_4$ will require=$\frac{1}{3}\times 2.55=0.85moles$  of $Al_2(SO_4)_3$

Mass of $Al_2(SO_4)_3=moles\times {\text {Molar mass}}=0.85moles\times 342g/mol=290g$

Thus 290 g of aluminium sulphate is produced.