Answer: 11.2 grams
Explanation:
The balanced chemical equation for reaction is:
[tex]BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl[/tex]
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
a) moles of [tex]BaCl_2[/tex]
[tex]\text{Number of moles}=\frac{10.0g}{208.23g/mol}=0.0480moles[/tex]
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 [tex]BaCl_2[/tex] produces= 1 mole of [tex]BaSO_4[/tex]
Thus 0.0480 moles of [tex]BaCl_2[/tex] require=[tex]\frac{1}{1}\times 0.0480=0.0480moles[/tex] of [tex]BaSO_4[/tex]
Mass of [tex]BaSO_4=moles\times {\text {Molar mas}}=0.0480mol\times 233.38g/mol=11.2g[/tex]
Thus 11.2 g of barium sulfate are produced if 10.0 grams of barium chloride are reacted is excess sulfuric acid
