Silver sulfate is precipitated from solution to the following reaction 2AgNO 3 (aq)+Na2SO 4 (aq) Ag 2 SO 4 (s)+2 NaNO 3 (aq)

Question

Silver sulfate is precipitated from solution to the following reaction 2AgNO 3 (aq)+Na2SO 4 (aq) Ag 2 SO 4 (s)+2 NaNO 3 (aq) If 3.80 moles of AgNO 3 and 2.79 moles of Na2SO4 present initially, how many moles of remain after precipitationis complete. Answer should be in 2 decimal places

Answer 1

First, you need to find the limiting reactant in this reaction. Which will run out first, given that for every mole of Na2SO4 we need two moles of AgNO3?

3.8 moles of AgNO3 can react with up to 1.9 moles of Na2SO4, and 2.79 moles of Na2SO4 could react with up to 5.58 moles of AgNO3. We don't have 5.58 moles of AgNO3, but we do have more than 1.9 moles of Na2SO4. So, Na2SO4 will be completely used up and is the limiting reactant.

From there, just look at how many moles of each reactant and product we will have. Every 2 moles of AgNO3 reacts with 1 mole of Na2SO4, producing one mole of Ag2SO4 and 2 moles of NaNO 3. Since our limiting reactant is 3.8 moles of AgNO3, we can use up 1.9 moles of Na2SO4, producing 1.9 moles of Ag2SO4 and 3.8 moles of NaNO3. What is left of the remaining reactant is 2.79 - 1.9 = 0.89 moles of Na2SO4.