Question

Al(s) + H₂SO₄(aq) --> Al₂(SO₄)₃(aq) + H₂(g)
Consider the unbalanced equation above. What volume of 0.465 M H₂SO₄ is needed to react with excess aluminum to produce 3.94 g of Al₂(SO₄)₃? Use a molar mass with at least as many significant figures as the data given.

Answer 1

2 Al(s) + 3 H2SO4(aq) --> Al2(SO4)3(aq) + 3 H2(g)

Molar mass of Al2(SO4)3,

MM = 2*MM(Al) + 3*MM(S) + 12*MM(O)

= 2*26.98 + 3*32.07 + 12*16.0

= 342.17 g/mol

mass(Al2(SO4)3)= 3.94 g

number of mol of Al2(SO4)3,

n = mass of Al2(SO4)3/molar mass of Al2(SO4)3

=(3.94 g)/(342.17 g/mol)

= 1.151*10^-2 mol

from balanced reaction above,

moles of H2SO4 required = 3*moles of Al2(SO4)3 formed

= 3*1.151*10^-2 mol

= 0.03453 mol

Now use:

mol of H2SO4 = M(H2SO4)*V(H2SO4)

0.03453 mol = 0.465 M * V(H2SO4)

V(H2SO4) = 0.0743 L

V(H2SO4) = 74.3 mL

Answer: 74.3 mL