## Chemistry 12th Edition

a. $NH_{4}NO_{3}$ -> $N_{2}O+2H_{2}O$ b. $20g$ $N_{2}O$
a. $NH_{4}NO_{3}$ -> $N_{2}O+2H_{2}O$ b. mass of $N_{2}O$ = $\frac{(0.46molNH_{4}NO_{3})\times(1molN_{2}O)\times(44.02gN_{2}O)}{(1molNH_{4}NO_{3})\times(1molN_{2}O)}$ $=20g$ $N_{2}O$