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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