Answer to Question #97474 in General Chemistry for Christopher Garner

Question #97474
could use some help finding correct answer - i know how to balance it...just can't get the correct answer

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.10 g of sodium carbonate is mixed with one containing 4.430 g of silver nitrate. How many grams of sodium carbonate, silver nitrate, silver carbonate, and sodium nitrate are present after the reaction is complete
1
Expert's answer
2019-10-29T06:44:33-0400

The chemical equation that occurs between reactants is following:

2AgNO3 + Na2CO3 = Ag2CO3 + 2NaNO3

First, we need to calculate the amount of each reagent in moles.

"n(Na_2CO_3)=\\frac{m(Na_2CO_3)}{M(Na_2CO_3)}=\\frac{3.10g}{106\\frac{g}{mol}}=0.0292 mol"

"n(AgNO3)=\\frac{m(AgNO3)}{M(AgNO3)}=\\frac{4.430g}{170\\frac{g}{mol}}=0.0261 mol"

According to stoichiometry of the reaction silver nitrate reacts with sodium carbonate in a ratio 2:1. Comparing with actual amount of reagents it is clear that silver nitrate is limiting reagent and sodium carbonate is in excess.

Thus, amount of Na2CO3 involved in the reaction is:

"n(react. Na_2CO_3)=\\frac{1}{2} \\cdot n(AgNO_3)=0.01305 mol"

"m(react. Na_2CO_3)=n(react. Na_2CO_3) \\cdot M(react. Na_2CO_3)=0.01305 mol \\cdot 106\\frac{g}{mol}=1.38g"

"m(remain. Na_2CO_3)=m(Na_2CO_3)-m(react. Na_2CO_3)=3.10g-1.38g=1.72g"

The amount of silver carbonate formed is:

"n(Ag_2CO_3)=\\frac{1}{2} \\cdot n(AgNO_3)=0.01305 mol"

"m(Ag_2CO_3)=n(Ag_2CO_3) \\cdot M(Ag_2CO_3)=0.01305 mol \\cdot 276\\frac{g}{mol}=3.60g"

The amount in moles of sodium nitrate is the same as silver nitrate:

"n(NaNO_3)=n(AgNO_3)=0.0261 mol"

"m(NaNO_3)=n(NaNO_3) \\cdot M(NaNO_3)=0.0261mol \\cdot 85 \\frac{g}{mol}=2.22 g"


Thus during the reaction silver nitrate is consumed completely and the following products are in the mixture:

AgNO3 - 0.00g

Na2CO3 - 1.72g

Ag2CO3 - 3.60g

NaNO3 - 2.22g


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