Question

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?

Answer 1

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

Step-by-step 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