Limiting Reactant-Excess Reactant - Sodium Carbonate And Silver Nitrate Reaction
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Limiting Reactant-Excess Reactant - Sodium Carbonate And Silver Nitrate Reaction Solutions of sodium carbonate and silver nitrate react as follows:

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

A solution containing 3.50 g of sodium carbonate is mixed with a solution containing 5 g of silver nitrate.
How many grams of sodium carbonate, silver nitrate, silver carbonate and sodium nitrate are present after the reaction is complete?

The strings: S7P6A66 (interaction - chemical).
S7P1A15 (containership - mass).

The math:
Although we are calculating, mass, grams and moles, the chemical reaction is the reason for the calculation. So Pj Problem of Interest is of type interaction (chemical). The amounts of products that can be produced in a chemical reaction, depend on the availability of the reactants in the reaction. Sometimes, the mole-ratio of reactants in a chemical reaction is less that the mole-ratio indicated in the balanced chemical equation of the reaction. The reactant that does not meet its mole-ratio requirement is the limiting reactant. The excess reactant is the amount of reactant left over after the limiting reactant has been used up. Once the limiting reactant is used up, the reaction will stop and no more products will be produced.

Na2CO3(aq) + 2AgNO3(aq) -------> Ag2CO3(s) + 2NaNO3(aq)
Mole-ratio from balanced equation = 1 mole of Na2CO3:2 moles of AgNO3

Molar mass of Na2CO3 = 46 + 12 + 48 = 106 g
So, 3.5 g of Na2CO3 = 3.5/106 = 0.033 mol

Molar mass of AgNO3 = 108 + 14 + 48 = 170 g
So, 5 g of AgNO3 = 5/170 = 0.0295 mol
Comparison of mole ratios indicate that AgNO3 is the limiting reactant
So, there is 0 g of AgNO3 left after the reaction is complete.

2 moles of AgNO3 reacts with 1 mole of Na2CO3
So, 0.0295 of AgNO3 will react with 0.0295/2 = 0.1475 mol of Na2CO3
Molar mass of Na2CO3 = 106 g
So, 0.01475 mol of Na2CO3 = 106(0.01475) = 1.56 g
So, 3.50 - 1.56 g = 1.94 g of Na2CO3 is left after the reaction is complete

From balanced equation, 2 moles of AgNO3 produced 1 mole Ag2CO3
So, 0.0295 mol of AgNO3 will produce 0.0295/2 = 0.01475 mol of Ag2CO3
Molar mass of Ag2CO3 = 216 + 12 + 48 = 276 g
So, 0.01475 mol of Ag2CO3 = 276(0.01475) = 4.07 g
So, 4.07 g of Ag2CO3 is produced after the reaction is complete.

From balanced equation, 2 moles of AgNO3 produced 2 moles NaNO3
So, 0.0295 mol of AgNO3 will produce 0.0295 mol of NaNO3
Molar mass of NaNO3 = 23 + 14 + 48 = 85 g
So, 0.0295 mol of NaNO3 = 85(0.0295) = 2.5 g
So, 2.5 g of NaNO3 is produced after the reaction is complete.

Math

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