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

Step-by-step 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}=(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=
`(1)/(1)* 0.0480=0.0480moles` of
`BaSO_4`

Mass of
`BaSO_4=moles* {\text {Molar mas}}=0.0480mol* 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​