In the reaction
` x Mg + yHNO_3 rarr Mg (NO_(3))_(2) + N_2O+ H_2 O`.

Options

A. ` x=2 , y= 5`
B. ` x =4 , y=10`
C. `x=3 , y=8`
D. ` x=5, y=9`

Answer

B

Explanation

.
Sice there are two N atoms in ` N_2 O`, we must calculate the change in oxdation number for two N atoms, i.e., multiply ` HNO_3` by `2`. Thus, total decrease in the oxidation number of N is 8 units. To equlize increase and decrease of oxidation number, we multiply Mg on both sides by 4 :
` 4 Mg + 2 HNO_3 rarr 4Mg(NO_3)_2 + N_2O +H_2 O`
To balance N (2 on RHS and ` 10` on LHS),we change the coefficient of `HNO_(3)` from 2 to 10 :
`4Mg(s)+10HNO_(3) rarr 4Mg(NO_(3))_(2)+N_(2)O+H_(2)O`
To balance O (30 on RHS and 26 on LHS), change the coefficient of `H_(2)O` from 1 to 5. Note that H atoms get automatically balanced:
`4Mg+10HNO_(3) rarr 4Mg(NO_(3))_(2)+N_(2)O+5H_(2)O`