Question

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 7.00 g of sodium carbonate is mixed with one containing 4.75 g of silver nitrate. After the reaction is complete, the solutions are evaporated to dryness, leaving a mixture of salts. How many grams of each of the following compounds are present after the reaction is complete?

sodium carbonate

silver nitrate

silver carbonate

sodium nitrate

Answer 1

no of moles of Na2CO3   = W/G.M.Wt

                                      = 7/106   = 0.066moloes

no of moles of AgNO3   = W/G.M.Wt

                                    = 4.75/169.87     = 0.028moles

Na2CO3(aq) + 2AgNO3(aq) -------------> Ag2CO3(s) + 2NaNO3(aq)

1 mole of Na2CO3 react with 2 moles of AgNO3

0.066moles of Na2Co3 react with = 2*0.066/1   = 0.132moles of AgNO3 is required

AgNO3 is limiting reactant

2 moles of AgNO3 react with excess of Na2CO3 to gives 1 mole of Ag2CO3

0.028moles of AgNO3 react with excess of Na2CO3 to gives = 1*0.028/2   = 0.014 moles of Ag2CO3

mass of Ag2CO3 = no of moles * gram molar mass

                            = 0.014*275.7453   = 3.86g

2 moles of AgNO3 react with excess of Na2CO3 to gives 2 mole of NaNO3

0.028moles of AgNO3 react with excess of Na2CO3 to gives   = 2*0.028/2 = 0.028 moles of NaNO3

mass of NaNO3 = no of moles * gram molar mass

                          = 0.028*84.9947

                          = 2.38g

2 moles of AgNO3 react with 1 mole of Na2CO3

0.028 moles of AgNO3 react with = 1*0.028/2   = 0.014moles of Na2Co3

Na2CO3 is excess reactant

The no of moles of Na2CO3 left after complete the reaction = 0.066-0.014   = 0.052 moles of Na2CO3

mass of Na2CO3 = no of moles * gram molar mass

                           = 0.052*106   = 5.512g

The mass of AgNO3 is 0g after complete the reaction because of it is limiting reactant