Question

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.40 g of sodium carbonate is mixed with one containing 4.86g of silver nitrate.
Required:
How many grams of sodium carbonate, silver nitrate, silver carbonate, and sodium nitrate are present after the reaction is complete?

Answer 1

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:

$AgNO_3~+~Na_2CO_3~->~Ag_2CO_3~+~NaNO_3$

Now, we can balance the reaction:

$2AgNO_3~+~Na_2CO_3~->~Ag_2CO_3~+~2NaNO_3$

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

$3.40~g~Na_2CO_3\frac{105.98~g~Na_2CO_3}{1~mol~Na_2CO_3}=0.032~mol~Na_2CO_3$

$4.86~g~AgNO_3\frac{169.8~g~AgNO_3}{1~mol~AgNO_3}=0.0286~mol~AgNO_3$

Divide by the coefficient

$\frac{0.032~mol~Na_2CO_3}{1}=0.032$

$\frac{0.0286~mol~AgNO_3}{2}=0.0143$

The smallest value is for $AgNO_3$ , therefore the 4.86 g of $AgNO_3$ .

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

$0.0286~mol~AgNO_3\frac{1~mol~AgCO_3}{2~mol~AgNO_3}\frac{275.74~g~AgCO_3}{1~mol~AgCO_3}=3.94~g~AgCO_3$

Amount of Sodium Nitrate

$0.0286~mol~AgNO_3\frac{2~mol~NaNO_3}{2~mol~AgNO_3}\frac{84.99~g~NaNO_3}{1~mol~NaNO_3}=2.43~g~NaNO_3$

Amount of Sodium Carbonate (Excess reactive)

$0.0286~mol~AgNO_3\frac{1~mol~NaCO_3}{2~mol~AgNO_3}\frac{105.98~g~NaCO_3}{1~mol~NaCO_3}=1.51~g~NaCO_3$

$3.4~g~NaCO_3-1.51~g~NaCO_3=1.89~g~NaCO_3$

Amount of Silver Nitrate

All the silver nitrate would be consumed in the reaction

I hope it helps!