Question

Aqueous solution of sodium carbonate and silver (I) nitrate react to form solid silver (I) carbonate and a solution of sodium mitrate. A solution containing 3.50g of sodium carbonate is mixed with 100.mL of 0.294 M silver nitrate. Write the balanced chemical equation. Write the net ionic equation.Phases must be included. How manyy grams of sodium carbonate, silver nitrate, silver carbonate and sodium nitrate are present after the reaction is complete ? Could you write down the step by step process for me please ?! Thank You.

Answer 1

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

2Na^+(aq) +CO3^2-(aq) + 2Ag^+(aq) + 2NO3^-(aq) ------> Ag2CO3(s) + 2Na^+(aq) + 2NO3^-(aq)

CO3^2-(aq) + 2Ag^+(aq) ------> Ag2CO3(s)   net ionic equation

no of moles of Na2CO3   = W/G.M.Wt

                                        = 3.5/106   = 0.033 moles

no of moles of AgNO3   = molarity * volume in L

                                     = 0.294*0.1 = 0.0294 moles

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

1 moles of Na2CO3 react with 2 moles of AgNO3

0.033 moles of Na2CO3 react with = 2*0.033/1 = 0.066 moles of AgNO3

AgNO3 is limiting reactant

2 moles of AgNO3 react with excess of Na2Co3 to gives 1 moles of Ag2Co3

0.0294 moles of AgNO3 react with excess of Na2CO3 to gives = 1*0.0294/2= 0.0147 moles of Ag2Co3

mass of Ag2CO3 = no of moles * gram molar mass

                            = 0.0147*275.7453   = 4.053g

2 moles of AgNO3 react with excess of Na2Co3 to gives 2 moles of NaNO3

0.0294 moles of AgNO3 react with excess of Na2CO3 to gives   =2*0.0294/2= 0.0294 moles of NaNO3

mass of NaNO3 = no ofmoles * gram molar mass

                             = 0.0294*84.9947   = 2.5g

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

2 moles of AgNO3 react with 1 mole of Na2CO3

0.0294 moles of AgNO3 react with = 1*0.0294/2   = 0.0147 moles of Na2CO3 is required

Na2CO3 is excess reactant

The no of moles of excess reactant remains after complete the reaction =0.033-0.0147 = 0.0183 moles

The amount of excess reactant remains after complete the reaction = 0.0183*106   = 1.94g

AgNO3 is limiting reactant.So it is complete the cumsume in the reaction. After complete the reaction the mass of AgNO3 is = 0 g