Answer to Question #133876 in General Chemistry for Jay

Question

A chemist adds 226 mL of a solution of Na2SO4 to 500 mL of 1.308 M AgCH3CO2. Assume the reaction goes to 100% completion and that Na2SO4 is the limiting reagent. If 76.81 g of Ag2SO4 (mM = 311.802 g/mol) is produced, what was the concentration, in mol/L, of the Na2SO4 solution that was added

Answer 1

Na₂SO₄+2AgCH₃CO₂=Ag₂SO₄+2CH₃COONa

1.in the case condition the "limiting reagent" is Na₂SO₄. that is, Na₂SO₄ was completely reacted

2.Based on the above reaction, 1 mole of Na₂SO₄ produces 1 mole (311.8gr)of Ag₂SO₄.

3.We find how many moles of Na₂SO₄ were used to produce 76.81 g of Ag₂SO₄:

1mole Na₂SO₄— 311.8 gr Ag₂SO₄

x mole Na₂SO₄ — 76.81 gr Ag₂SO₄

x= 76.81*1/311.8=0.246 mole Na₂SO₄

4.We find the concentration of Na₂SO₄.To do this, we divide the amount of Na₂SO₄ (mol) by the volume (liters):

0.246 mol/ 0.226 litrs = 1.09 mol/L ≈1.1 mol/l

ANSWER: 1.1mol/L