Question

23. In the following reaction, how many grams of sodium sulfate are needed to react with 3.08 moles of strontium nitrate? (Hint: stoichiometry mol Na2SO4: mol Sr(NO3)2 = 1 mol: 1 mol) Sr(NO3)2 (aq) + Na2SO4 (aq) --> SrSO4 (s) + 2 NaNO3 (aq) (balanced?)

Answer 1

Sr(NO₃)₂ (aq) + Na₂SO₄ (aq)    SrSO₄ (s) + 2 NaNO₃ (aq)

Molar mass of Na₂SO₄ = 142.04 g/mol

Molar mass of Sr(NO₃)₂ = 211.63 g/mol

In this reaction,

1 mole of Na₂SO₄ reacts with 1 mole of Sr(NO₃)₂.

3.08 moles of Na₂SO₄ reacts with 3.08 moles of Sr(NO₃)₂.

Now, 1 mole of Na₂SO₄ = 142.04 g

3.08 moles of Na₂SO₄ = (3.08 x 142.04) g = 437.48 g

So, 437.48 g of sodium sulfate are needed to react with 3.08 moles of strontium nitrate.