Question

An excess of sodium carbonate, Na2CO3, in solution is added to a solution containing 15.71 g CaCl2. After performing the experiment, 13.19 g of calcium carbonate, CaCO3, is produced. Calculate the percent yield of this reaction

Answer 1

Answer:

93.15 %

Explanation:

We have to start with the chemical reaction:

$CaCl_2~+~Na_2CO_3~->~CaCO_3~+~NaCl$

Now, we can balance the reaction:

$CaCl_2~+~Na_2CO_3~->~CaCO_3~+~2NaCl$

Our initial data are the 15.71 g of $CaCl_2$, so we have to do the following steps:

1) Convert from grams to moles of $CaCl_2$ using the molar mass (110.98 g/mol).

2) Convert from moles of $CaCl_2$ to moles of $CaCO_3$ using the molar ratio. ( 1 mol $CaCl_2$= 1 mol of $CaCO_3$).

3) Convert from moles of $CaCO_3$ to grams of $CaCO_3$ using the molar mass. (100 g/mol).

$15.71~g~CaCl_2\frac{1~mol~CaCl_2}{110.98~g~CaCl_2}\frac{1~mol~CaCO_3}{1~mol~CaCl_2}\frac{100~g~CaCO_3}{1~mol~CaCO_3}=14.16~g~CaCO_3$

Finally, we can calculate the yield percent:

$%~=~\frac{13.19~g~CaCO_3}{14.16~g~CaCO_3}*100=93.15~%$

I hope it helps!