Answer:
H2 produced = 0.4235g
Explanation:
Equations for reaction with Hcl
Aluminium
2Al + 6 Hcl ------------ 2Alcl3 + 3H2
Magnesium
Mg + 2Hcl -------- Mgcl2 + H2
Aluminium = 70% of 0.710g of sample
sample contains (70 x 0.710)/100 = 0.497g of aluminium
Magnesium = 30% of 0.710g of sample
sample contains(30 x 0.710)/100 = 0 .213g of magnesium
Moles = mass/ molecular mass
Moles of aluminium in sample = 0.497/27 = 0.0184
2 moles of aluminium gives 3 moles of H2
No of moles of H2 from reaction with aluminium = (2 x0.0184)/3
= 0.0123 moles
1 mole of H2 = 2g therefore mass of H2 produced = 0.0123 x 2 = 0.0246g
Moles of magnesium in sample = 0.213/24 = 0.008875
1 mole of mg gives 1 mole of H2
No of moles of H2 from reaction with magnesium = 0.008875 x 1
= 0.008875
1 mole of H2 = 2g therefore mass of H2 produced = 0.008875 x 2
= 0.01775g
Ttal mass of H2 = 0.0246 +0.01775 = 0,04235g
