Question

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.00 g of sodium carbonate is mixed with one containing 4.75 g of silver nitrate. 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

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

Given that

3.00 g of sodium carbonate

4.75 g of silver nitrate.

Number of moles =- amount in g / molar mass

Amount in g = number of moles * molar mass

Limiting agent

The limiting reagent (or limiting reactant) is a substance that limits the amount of product in the reaction. This limiting agent is completely consumed in the reaction.

Excess reagent is the reagent which left after reaction.

mol of Na2CO3 = 3.00g / 105.99g/mol = 0.0283 mol

mol of AgNO3 = 4.75 g / 169.87 g/mol = 0.0280 mol AgNO3

limiting agent is AgNO3

mol of Na2CO3 reacted = 1/2 x 0.0280 = 0.014 mol

mol of Na2CO3 excess = 0.0283 - 0.0140 = 0.0143 mol

after reaction

AgNO3= 0 g

Na2CO3 in g = 0.0143 mol* 105.98844 g/mol

= 1.52 g

Moles of Ag2CO3= 0.0280 mol AgNO3 *1/2

=0.0143 mol Ag2CO3

Amount of Ag2CO3 =

0.0143 mol Ag2CO3* 275.7453 g/mol

= 3.94 g Ag2CO3

Mole of NaNO3 =0.0280 mol AgNO3 *1/1

= 0.0280 mol NaNO3

Amount in g =0.0280 mol NaNO3*84.99467 g/mol

=2.38 g NaNO3