Question

NH₄NO₃ → N₂O + 2H₂O When 45.70 g of NH₄NO₃ decomposes, what mass of each product is formed?

Answer 1

Answer: 25.13 g of
`N_2O` and 20.56 g of
`H_2O` will be produced from 45.70 g of
`NH_4NO_3`

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} NH_4NO_3=(45.70g)/(80.04g/mol)=0.571moles`

The balanced chemical equation is:

`NH_4NO_3\rightarrow N_2O+2H_2O`

According to stoichiometry :

1 mole of
`NH_4NO_3` produce = 1 mole of
`N_2O`

Thus 0.571 moles of
`NH_4NO_3` will require=
`(1)/(1)* 0.571=0.571moles` of
`N_2O`

Mass of
`N_2O=moles* {\text {Molar mass}}=0.571moles* 44.01g/mol=25.13g`

1 mole of
`NH_4NO_3` produce = 2 moles of
`H_2O`

Thus 0.571 moles of
`NH_4NO_3` will require=
`(2)/(1)* 0.571=1.142moles` of
`H_2O`

Mass of
`H_2O=moles* {\text {Molar mass}}=1.142moles* 18g/mol=20.56g`

Thus 25.13 g of
`N_2O` and 20.56 g of
`H_2O` will be produced from 45.70 g of
`NH_4NO_3`