Question

For the following reaction, 67.1 grams of barium hydroxide are allowed to react with 36.3 grams of sulfuric acid. barium hydroxide (aq) sulfuric acid (aq) barium sulfate (s) water (l) What is the maximum amount of barium sulfate that can be formed

Answer 1

Answer: 86.2 g of
`BaSO_4` will be produced from the given masses of both reactants.

Step-by-step explanation:

To calculate the moles, we use the equation:

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

a) moles of
`Ba(OH)_2`

`\text{Number of moles}=(67.1g)/(171g/mol)=0.39moles`

b) moles of
`H_2SO_4`

`\text{Number of moles}=(36.3g)/(98g/mol)=0.37moles`

`Ba(OH)_2+H_2SO_4\rightarrow BaSO_4+2H_2O`

According to stoichiometry :

1 moles of
`H_2SO_4` require 1 mole of
`Ba(OH)_2`

Thus 0.37 moles of
`H_2SO_4` require=
`(1)/(1)* 0.37=0.37moles` of
`Ba(OH)_2`

Thus
`H_2SO_4` is the limiting reagent as it limits the formation of product.

As 1 mole of
`H_2SO_4` give = 1 mole of [texBaSO_4[/tex]

Thus 0.37 moles of
`H_2SO_4` give =
`(1)/(1)* 0.37=0.37moles` of
`BaSO_4`

Mass of
`BaSO_4=moles* {\text {Molar mass}}=0.37moles* 233g/mol=86.2g`

Thus 86.2 g of
`BaSO_4` will be produced from the given masses of both reactants.