Question

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

Answer 1

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