Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 7.50 g of sodium carbonate is mixed with one containing 6.75 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?

After the reaction we have 5.46 grams of Ag2CO3, 3.37 grams of NaNO3 and remain 5.40 grams of Na2CO3

Explanation:

Step 1: Data given

A solution containing 7.50 g of sodium carbonate is mixed with one containing 6.75 g of silver nitrate.

Step 2: The balanced equation

Na2CO3 + 2 AgNO3 → Ag2CO3 + 2 NaNO3

Step 3: Calculate moles

Moles = mass / molar mass

moles Na2CO3 = 7.50 g/105.99 g/mol=0.0708 moles

moles AgNO3 = 6.75 g / 169.87 g/mol = 0.0397 moles (limiting reactant)

Step 4: Calculate limiting reactant

For 1 mol Na2CO3 we need 2 moles AgNO3 to produce 1 mol Ag2CO3 and 2 moles NaNO3

AgNO3 is the limiting reactatn. It will completely be consumed. (0.0397 moles). Na2CO3 is in excess. There will react 0.0397/2 = 0.01985 moles

There will remain 0.0708 - 0.01985 = 0.05095 moles Na2CO3

This is 0.05095 moles * 105.99 g/mol = 5.40 grams

Step 5: Calculate moles products

For 1 mol Na2CO3 we need 2 moles AgNO3 to produce 1 mol Ag2CO3 and 2 moles NaNO3

For 0.0397 moles AgNO3 we'll have 0.0397 moles NaNO3 and 0.01985 moles Ag2CO3

Step 6: Calculate mass

Mass Ag2CO3 = moles Ag2CO3 * molar mass

Mass Ag2CO3 = 0.01985 moles * 275.75 g/mol

Mass Ag2CO3 = 5.46 grams

Mass NaNO3 = 0.0397 moles * 84.99 g/mol

Mass NaNO3 = 3.37 grams

After the reaction we have 5.46 grams of Ag2CO3, 3.37 grams of NaNO3 and remain 5.40 grams of Na2CO3

7.50 + 6.75 grams = 14.25

Mass products = 5.46 + 3.37 grams = 8.83

mass of reactants in excess = 5.40 grams

remaining products = 8.83 + 5.40 = 14.23 grams