Question

250 ml of 0.5M Na2SO4 solution are added to an aqueous solution containing 10g of BaCl2 resulting in the formation of white precipitate of BaSO4. How many moles and how many grams of BaSO4 will be obtained? ((Molar mass of BaSO4 = 233, BaCl2 = 208))

Answer 1

Moles of BaSO4=0.048
Weight of BaSO4=11.20g

Answer 2

Answer: 0.048 moles , 11.20 grams

Explanation:

$Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl$

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{moles}{\text {Volume in L}}$

moles of $Na_2SO_4=Molarity\times {\text {Volume in L}}=0.5\times 0.25=0.125moles$

$\text{Number of moles}of BaCl_2=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{10g}{208g/mol}=0.048moles$

According to stoichiometry:

1 mole of  $Na_2SO_4$ combines with 1 mole of $BaCl_2$ to give 1 mole of $BaSO_4$

0.048 moles of $Na_2SO_4$ combines with =$\frac{1}{1}\times 0.048=0.048moles$  of $BaCl_2$ to give 0.048 mole of $BaSO_4$

Thus $Na_2SO_4$ is the limiting reagent as it limits the formation of product.

0.048 moles of $BaCl_2$ produces= 0.048 moles of $BaSO_4$

Mass of $BaSO_4=moles\times {\text{Molar mass}}=0.048moles\times 233g/mol=11.20g$

Thus 0.048 moles of $BaSO_4$ and 11.20 grams of $BaSO_4$ is produced.