Question

A 0.250-g sample of a magnesium-aluminum alloy dissolves completely in an excess of HCl(aq). When the liberated H2(g) is collected over water at 29 ∘C and 752 torr, the volume is found to be 345 mL . The vapor pressure of water at 29 ∘C is 30.0 torr. A) How many moles of H2 can be produced from x grams of Mg in magnesium-aluminum alloy? The molar mass of Mg is 24.31 g/mol. B) How many moles of H2 can be produced from y grams of Al in magnesium-aluminum alloy? The molar mass of Al is 26.98 g/mol.

Answer 1

rxn of Mg and Al with HCl:

Mg(s)+2HCl(aq)--->H2(g)+MgCl2(aq)

2Al(s)+6HCl(aq)--->3H2(g) +2AlCl3(aq)

Let mol of H2 produced from 0.250 g alloy sample=n

Using ideal gas equation,

n=PV/RT

where P=pressure of H2=752 torr-vp of water=752-30=722 torr=722 torr*(0.00131579 atm/torr)=0.95atm

V=345ml=0.345L

R=universal gas constant=0.0821 L atm/Kmol

T=29+273=302K

n=0.95atm*0.345L/(0.0821Latm/Kmol)*302K=0.0132 mol

A) mol of Mg=xg/24.31g/mol=x/24.31 mol

So ,as 1 mol Mg yields 1 mol H2 ,so mol of H2 formed=x/24.31 mol=0.0411x mol

B) mol of Al =yg/26.98g/mol=y/26.98 mol

As 2 mol of Al produces 3 mol of H2,

So,y/26.98 mol of Al produces H2=(y/26.98)mol Al*(3molH2/2molAl)=0.0556y mol

given y+x=0.250 g............(1)

And total mol of H2 produced=0.0556 y+0.0411x=0.0132...........(2)

putting the value of x from (1) into (2),

0.0556y+0.0411(0.250-y)=0.0132

0.0145y=0.00292

y=0.201g

x=0.250-0.201=0.049g