How many grams of each of the following compounds are present after the reaction is complete?

Question

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 4.25 g of sodium carbonate is mixed with one containing 7.50 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

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate

 

Na2CO3(aq) + 2AgNO3(aq) → Ag2CO3(s) + 2NaNO3(aq)

 

this is the balanced equation.

 

4.25 g of sodium carbonate x (1 mole Na2CO3/105.99 g Na2CO3) = .0401 moles Na2CO3

 

7.50 g of silver nitrate x (1 mole AgNO3/169.87 g AgNO3) = .0442 moles AgNO3

 

Since for every 1 of the Na2CO3, you need 2 of the AgNO3, all of the AgNO3 will react with .0221 moles of Na2CO3

 

So, so far, there will be no AgNO3 left and there will be .0180 moles of Na2CO3 left

 

.0180 moles of Na2CO3 x (105.99 g Na2CO3/1 mole Na2CO3) = 1.91 g Na2CO3  left

 

.0442 moles AgNO3 x (1 mol Ag2CO3/2 mol AgNO3) = 0.0221 mol Ag2CO3

 

0.0221 mol Ag2CO3 x (275.75 6.09 g Ag2CO3/1 mol Ag2CO3) =  6.09 g Ag2CO3

 

.0442 moles AgNO3 x (2 mol NaNO3/2 mol AgNO3) = .0442 mole NaNO3

 

This was very long to type.  You better check my math...