Calculate the number of oxygen atoms requried to combine with `7.0 g` of `N_(2)` to form `N_(2) O_(3)` if 82% of `N_(2)` is converted into products.
`N_(2) + (3)/(2) O_(2) rarr N_(2) O_(3)`

Options

A. `3.24 xx 10^(23)`
B. `3.6 xx 10^(23)`
C. `18 xx 10^(23)`
D. `6.02 xx 10^(23)`

Answer

B

Explanation

`2N_(2) + 3O_(2) rarr 2N_(2) O_(3)`, (%yield = 80%)
Initial mol of `N_(2) = (7.0)/(28) = 0.25 "mol"`
Mole of `N_(2)` converted `= 0.25 xx (80)/(100) = 0.2`
`2 "mo" N_(2) = 3 "mol" O_(2)`
`0.2 "mol" N_(2) = 0.3 "mol" O_(2) (1 "mol" O_(2) = 2` oxygen atom)
`= 2 xx 0.3 "mol" O "atom"`
`= 2 xx 0.3 xx 6.02 xx 10^(23) = 3.6 xx 10^(23)`