Question

If 355 mL of 1.50 M aluminum nitrate is added to an excess of sodium sulfate, how many grams of aluminum sulfate will be produced?

Answer 1

Answer is: 91.1 grams of aluminum sulfate.
Balanced chemical reaction: 2Al(NO₃)₃ + 3Na₂SO₄→ Al₂(SO₄)₃ + 6NaNO₃.
V(Al(NO₃)₃) = 355 mL ÷ 1000 mL/L = 0.355 L.
c(Al(NO₃)₃) = 1.5 mol/L.
n(Al(NO₃)₃) = V(Al(NO₃)₃) · c(Al(NO₃)₃).
n(Al(NO₃)₃) = 0,355 L · 1.5 mol/L.
n(Al(NO₃)₃) = 0.5325 mol.
From chemical reaction: n(Al(NO₃)₃) : n(Al₂(SO₄)₃) = 2 : 1.
n(Al₂(SO₄)₃) = 0.266 mol.
m(Al₂(SO₄)₃) = 0.266 mol · 342.15 g/mol.
m(Al₂(SO₄)₃) = 91.1 g.