Question

Sulfuric acid (H2SO4) dissolves Aluminum metal according to the reaction:
2 Al(s)+ 3 H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)
Suppose you want to dissolve an Aluminum block with a mass of 15.2 g. What minimum mass of H2SO4 (in g) do you need? What mass of H2 gas (in g) can the complete reaction of the aluminum block produce? Please Show Work

Answer 1

1)

Molar mass of Al = 26.98 g/mol

mass of Al = 15.2 g

mol of Al = (mass)/(molar mass)

= 15.2/26.98

= 0.5634 mol

According to balanced equation

mol of H2SO4 required = (3/2)* moles of Al

= (3/2)*0.5634

= 0.8451 mol

Molar mass of H2SO4,

MM = 2*MM(H) + 1*MM(S) + 4*MM(O)

= 2*1.008 + 1*32.07 + 4*16.0

= 98.086 g/mol

mass of H2SO4 = number of mol * molar mass

= 0.8451*98.09

= 82.89 g

Answer: 82.9 g

2)

According to balanced equation

mol of H2 formed = (3/2)* moles of Al

= (3/2)*0.5634

= 0.8451 mol

Molar mass of H2 = 2.016 g/mol

mass of H2 = number of mol * molar mass

= 0.8451*2.016

= 1.704 g

Answer: 1.70 g