Answer: 25.13 g of [tex]N_2O[/tex] and 20.56 g of [tex]H_2O[/tex] will be produced from 45.70 g of [tex]NH_4NO_3[/tex]
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} NH_4NO_3=\frac{45.70g}{80.04g/mol}=0.571moles[/tex]
The balanced chemical equation is:
[tex]NH_4NO_3\rightarrow N_2O+2H_2O[/tex]
According to stoichiometry :
1 mole of [tex]NH_4NO_3[/tex] produce = 1 mole of [tex]N_2O[/tex]
Thus 0.571 moles of [tex]NH_4NO_3[/tex] will require=[tex]\frac{1}{1}\times 0.571=0.571moles[/tex] of [tex]N_2O[/tex]
Mass of [tex]N_2O=moles\times {\text {Molar mass}}=0.571moles\times 44.01g/mol=25.13g[/tex]
1 mole of [tex]NH_4NO_3[/tex] produce = 2 moles of [tex]H_2O[/tex]
Thus 0.571 moles of [tex]NH_4NO_3[/tex] will require=[tex]\frac{2}{1}\times 0.571=1.142moles[/tex] of [tex]H_2O[/tex]
Mass of [tex]H_2O=moles\times {\text {Molar mass}}=1.142moles\times 18g/mol=20.56g[/tex]
Thus 25.13 g of [tex]N_2O[/tex] and 20.56 g of [tex]H_2O[/tex] will be produced from 45.70 g of [tex]NH_4NO_3[/tex]
