Question

How many grams of barium sulfate can be produced from the reaction of 2.54 grams sodium sulfate and 2.54 g barium chloride? Na2SO4(aq) + BaCl2(aq) --> BaSO4(s) + 2NaCl(aq) Report your answer to 3 decimal places.

Answer 1

Answer: 2.796 grams

Explanation:

$Na_2SO_4+BaCl_2\rightarrow 2NaCl+BaSO_4$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of sodium sulphate}=\frac{2.54g}{142g/mol}=0.018moles$

$\text{Number of moles of barium chloride}=\frac{2.54g}{208g/mol}=0.012moles$

According to stoichiometry:

1 mole of $BaCl_2$ reacts with 1 mole of $Na_2SO_4$

0.012 moles of $BaCl_2$ will react with=$\frac{1}{1}\times 0.012=0.012moles$ of $Na_2SO_4$

Thus $BaCl_2$ is the limiting reagent as it limits the formation of product. $Na_2SO_4$  is the excess reagent as (0.018-0.012)=0.006 moles are left unused.

1 mole of $BaCl_2$ produces 1 mole of $BaSO_4$

0.012 moles of $BaCl_2$ will produce=$\frac{1}{1}\times 0.012=0.012moles$ of $BaSO_4$

Mass of $BaSO_4=moles\times {\text {Molar mass}}=0.012\times 233=2.796g$

Thus 2.796 grams of $BaSO_4$  are produced.