consider the following reaction equation: 2Al(OH)\(_3\) + 3H\(_2\)SO\(_4\) →Al\(_2\)(SO4)\(_3\)(aq)+6H\(_2\)O. How many moles of tetraoxosulphate(vi)acid is needed to produce 0.5mol of aluminium tetraoxosulphate(vi)?
- 0.5
- 1.0
-
1.5
- 3.0
The correct answer is: C
Explanation
Where 3mols of H2So4 = 1 mol of (Al2SO4)3
y mols of H2SO4 = 0.5 mol of (Al2SO4)3
Then cross multiply to have:
y = \(\frac{3×0.5}{1}\)
y = 1.5 moles