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)

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

the millers tale is the answer.

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons