Question

250 g H 2 SO 4 completely reacted with aluminum?

2AI(s) + 32504(aq)- AI2 (504) 3(aq) + 3H 2 (9)
0.85g
450g
290g
870g

Answer 1

Answer: 290 g of aluminium sulphate is produced.

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} H_2SO_4=(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=
`(1)/(3)* 2.55=0.85moles` of
`Al_2(SO_4)_3`

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

Thus 290 g of aluminium sulphate is produced.