Question

If 1.08 g of sodium sulfate reacts with an excess of phosphoric acid, how much sulfuric acid is produced?

Answer 1

Answer: 0.745 g of
`H_2SO_4` will be produced from 1.08 g of sodium sulfate

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} Na_2SO_4=(1.08g)/(142.04g/mol)=0.0076moles`

`3Na_2SO_4+2H_3PO_4\rightarrow 2Na_3PO_4+3H_2SO_4`

`Na_2SO_4` is the limiting reagent as it limits the formation of product and
`H_3PO_4` is the excess reagent.

According to stoichiometry :

3 moles of
`Na_2SO_4` produce = 3 moles of
`H_2SO_4`

Thus 0.0076 moles of
`Na_2SO_4` will require=
`(3)/(3)* 0.0076=0.0076moles` of
`H_2SO_4`

Mass of
`H_2SO_4=moles* {\text {Molar mass}}=0.0076moles* 98.1g/mol=0.745g`

Thus 0.745 g of
`H_2SO_4` will be produced from 1.08 g of sodium sulfate