Respuesta :
Answer:
1.89 of Sodium Carbonate
3.94 g of Silver Carbonate
2.43 g of Sodium Nitrate
Zero grams of Silver Nitrate
Explanation:
We have to start with the reaction:
[tex]AgNO_3~+~Na_2CO_3~->~Ag_2CO_3~+~NaNO_3[/tex]
Now, we can balance the reaction:
[tex]2AgNO_3~+~Na_2CO_3~->~Ag_2CO_3~+~2NaNO_3[/tex]
Now, we have to calculate the limiting reagent and we have to follow a few steps:
1) Convert to moles (using the molar mass of each compound)
2) Divide by the coefficient of each reactive (given by the balanced reaction)
Convert to moles
[tex]3.40~g~Na_2CO_3\frac{105.98~g~Na_2CO_3}{1~mol~Na_2CO_3}=0.032~mol~Na_2CO_3[/tex]
[tex]4.86~g~AgNO_3\frac{169.8~g~AgNO_3}{1~mol~AgNO_3}=0.0286~mol~AgNO_3[/tex]
Divide by the coefficient
[tex]\frac{0.032~mol~Na_2CO_3}{1}=0.032[/tex]
[tex]\frac{0.0286~mol~AgNO_3}{2}=0.0143[/tex]
The smallest value is for [tex]AgNO_3[/tex] , therefore the 4.86 g of [tex]AgNO_3[/tex] .
Now we can calculate the amount of compounds produced is we follow a few steps:
1) Use the molar ratio
2) Convert to moles (using the molar mass of each compound)
Amount of Silver Carbonate
[tex]0.0286~mol~AgNO_3\frac{1~mol~AgCO_3}{2~mol~AgNO_3}\frac{275.74~g~AgCO_3}{1~mol~AgCO_3}=3.94~g~AgCO_3[/tex]
Amount of Sodium Nitrate
[tex]0.0286~mol~AgNO_3\frac{2~mol~NaNO_3}{2~mol~AgNO_3}\frac{84.99~g~NaNO_3}{1~mol~NaNO_3}=2.43~g~NaNO_3[/tex]
Amount of Sodium Carbonate (Excess reactive)
[tex]0.0286~mol~AgNO_3\frac{1~mol~NaCO_3}{2~mol~AgNO_3}\frac{105.98~g~NaCO_3}{1~mol~NaCO_3}=1.51~g~NaCO_3[/tex]
[tex]3.4~g~NaCO_3-1.51~g~NaCO_3=1.89~g~NaCO_3[/tex]
Amount of Silver Nitrate
All the silver nitrate would be consumed in the reaction
I hope it helps!
