Percent yield of chemical reactions

Question

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If 2.15g of water is produced from the reaction of 34.3g of sulfuric acid and 12.9g of sodium hydroxide, calculate the percent yield of water.

Be sure your answer has the correct number of significant digits in it.

Answer 1

Percent yield is a function of actual yield over maximum yield. To find it we need to know what the maximum potential yield is.

First, we need a balanced equation

[?]H₂SO₄(aq) + [?]NaOH(s) → [?]Na₂SO₄(aq) + [?]H₂O(l)

We have twice the amount of SO₄ that we need per Na, so doubling the NaOH will do the trick

H₂SO₄(aq) + 2 NaOH(s) → Na₂SO₄(aq) + 2 H₂O(l)

Second, we need to find molar quantity of our reaction

H₂SO₄= 98.079 g/mol

(34.3g H₂SO₄)/ (98.079 g/mol) = .350 mol

NaOH = 39.997 g/mol

(12.9g NaOH) / (39.997 g/mol) = .323 mol

Third, fill in a RICE table to find a limiting reaction and final products

Reaction: H₂SO₄(aq) + 2 NaOH(s) → Na₂SO₄(aq) + 2 H₂O(l)

Initial quantity: ____ .350 mol____.323 mol______________________

Change in Quantity: _-x __________-2x________+x__________+2x_ (the maximum of x is .323/2 or .162mol

Equilibrium:_______.188 mol____.000 mol____.162 mol_____.324 mol

NaOH is our limiting reactant. Full consumption of NaOH produces .324 mol of water

Finally, calculate expected mass of water and find percent yield

(.324 mo H₂O)*(18.02g/mol) = 5.84 g H₂O

% yield = 2.15g (actual H₂O)/ 5.84 g (expected H₂O) X 100 = 36.8% yield