Question

Magnalium alloys contain 70% Al and 30.0% Mg by mass. How many grams of H2(g) are produced in the reaction of a 0.710g sample of this alloy with excess HCl(aq)

Answer 1

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