Question

how many grams of barium sulfate are produced if 10.0 grams of barium chloride are reacted is excess sulfuric acid​

Answer 1

Answer: 11.2 grams

Explanation:

The balanced chemical equation for reaction is:

$BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl$

To calculate the moles, we use the equation:

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

a) moles of $BaCl_2$

$\text{Number of moles}=\frac{10.0g}{208.23g/mol}=0.0480moles$

According to stoichiometry :

As sulfuric acid​ is in excess , the limiting reagent is barium chloride as it limits the formation of product.

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

Thus 0.0480 moles of $BaCl_2$ require=$\frac{1}{1}\times 0.0480=0.0480moles$  of $BaSO_4$

Mass of $BaSO_4=moles\times {\text {Molar mas}}=0.0480mol\times 233.38g/mol=11.2g$

Thus 11.2 g of barium sulfate are produced if 10.0 grams of barium chloride are reacted is excess sulfuric acid​