Question

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.80 g of sodium carbonate is mixed with one containing 5.43 g of silver nitrate.
Part A)
How many grams of sodium carbonate are present after the reaction is complete?

Part B)
How many grams of silver nitrate are present after the reaction is complete?

Part C)
How many grams of silver carbonate are present after the reaction is complete?

Part D)
How many grams of sodium nitrate are present after the reaction is complete?

Answer 1

Answer:

Part A) $m_{Na_2CO_3}^{leftover}=3.74g$

Part B) Nothing as it is the liming reactant.

Part C) $m_{Ag_2CO_3}=4.42gAg_2CO_3$

Part D) $m_{NaNO_3} =2.72gNaNO_3$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$Na_2CO_3(aq)+2AgNO_3(aq)\rightarrow Ag_2CO_3(s)+2NaNO_3(aq)$

Thus, in order to proceed, we first need to identify the limiting reactant is it yielding the least moles of silver carbonate for instance, thus, for the given mass of sodium carbonate (molar mass 106 g/mol) and silver nitrate (molar mass 170 g/mol) we obtain for silver carbonate:

$n_{Ag_2CO_3}^{by\ Na_2CO_3}=3.80gNa_2CO_3*\frac{1molNa_2CO_3}{106gNa_2CO_3}*\frac{1molAg_2CO_3}{1molNa_2CO_3} =0.0359molAg_2CO_3\\\\n_{Ag_2CO_3}^{by\ AgNO_3}=5.43gAgNO_3*\frac{1molAgNO_3}{170gAgNO_3}*\frac{1molAg_2CO_3}{2molAgNO_3} =0.0160molAg_2CO_3$

In such a way, we can see that the silver nitrate yields less amount of silver carbonate, that is why it is the limiting reactant. Thus, we answer:

Part A) Since sodium carbonate is the excess reactant, we need to compute the consumed grams by the limiting silver nitrate as shown below:

$m_{Na_2CO_3}^{consumed}=5.43gAgNO_3*\frac{1molAgNO_3}{170gAgNO_3}*\frac{1molNa_2CO_3}{2molAgNO_3} *\frac{106gNa_2CO_3}{1molNa_2CO_3} =1.69gNa_2CO_3$

Therefore, the leftover of sodium carbonate is:

$m_{Na_2CO_3}^{leftover}=5.43g-1.69g=3.74g$

Part B) Since silver nitrate is the limiting reactant no leftover are present after the reaction as it was all consumed.

Part C) As the 5.43 g of silver nitrate yield 0.0160 moles of silver carbonate, the corresponding mass is computed as shown below:

$m_{Ag_2CO_3}=0.0160molAg_2CO_3*\frac{276gAg_2CO_3}{1molAg_2CO_3}=4.42gAg_2CO_3$

Part D) Finally, via stoichiometry, we compute the grams of sodium nitrate that are yielded:

$m_{NaNO_3}=5.43gAgNO_3*\frac{1molAgNO_3}{170gAgNO_3}*\frac{2molNaNO_3}{2molAgNO_3} *\frac{85gNaNO_3}{1molNaNO_3} =2.72gNaNO_3$

Best regards.