Question

What mass of Al2(SO4)3 will be produced from 355 ml of 0.200 M H2SO4 solution 2.86 g of Al

Answer 1

8.10 g

Step-by-step explanation:

3H2SO4 + 2Al --> Al2(SO4)3 + 3H2

Find the limiting reactant using stoichiometry

`\frac{.355 L H2SO4}{} ((.200 mol H2SO4)/(1L) )((1 mol Al2(SO4)3)/(3 mol H2SO4) )((342.14 g)/(1 mol Al2(SO4)3) ) = 8.10 g Al H2(SO4)3`

`\frac{2.86 g Al}{}((1 mol Al)/(26.98 g) )((1 mol Al2(SO4)3)/(2 mol Al) )((342.14 g)/(1 mol Al2(SO4)3) ) = 18.1 g Al2(SO4)3`

(Using 342.14 g/mol as molar mass of Al2(SO4)3)

8.10 < 18.1, so Al is the limiting reactant and 8.10 g of Al2(SO4)3 is made