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

Submitted by balesofhay on Tue, 08/23/2011 - 18:20

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 2.50 g of silver nitrate. How many grams of each of the following compounds are present after the reaction is complete? 

The only one i cannot find the answer to is sodium carbonate? Please Help? :)


1_ The first step would be to write a balanced equation. Assuming you used anhydrous sodium carbonate, you could write:

Na2CO3 + 2AgNO3 → Ag2CO3 + 2NaNO3

(You could also add "(s)" after Ag2CO3 and "(aq)" after all the other formulas, but that does not affect the stoichiometry).

The balanced reaction tells you that 1 mol Na2CO3 reacts with 2 moles AgNO3, or that 1 mol AgNO3 reacts with 0.5 moles Na2CO3 .

2_ The next step would be to find or calculate molecular weights/formula weights.

You would end up with something like:

105.99 g/mol for Na2CO3

169.88 g/mol for AgNO3, and so on.

3_ After that, you would want to find how many moles of each reactant you have. You could calculate

(3.00 g Na2CO3)/(105.99 g/mol ) = 0.0283 moles Na2CO3

(2.50 g AgNO3)/(169.88 g/mol) = 0.0147 moles AgNO3

4_ Next, you want to know if one reactant is in excess. If so, the other reactant is the limiting reactant. In that case you want to know the moles of limiting reactant and the excess (in moles) of the other reactant.

The moles of AgNO3 you have (0.0147) will react with 0.0074 moles of Na2CO3. Then you will have used up all the AgNO3. At that point, the reaction is complete. You will have a lot of unreacted Na2CO3. AgNO3 is your limiting reagent. You have excesss of Na2CO3. The amounts of products that will form depend on those 0.0147 moles of AgNO3, which will result in 0.0147 moles of NaNO3 and 0.074 moles of Ag2CO3. The excess Na2CO3 that you will have, unreacted, after the reaction is complete is

0.0283 moles - 0.0074 moles = 0.0209 moles

5_ The last step is calculating grams of each compound from the numbers of moles calculated.