Calculate the number of oxygen atoms required to combine with `7.0g" of "N_(2)` to from `N_(2)O_(3)` if 80% of `N_2` is converted into products.
`N_(2) + (3)/(2)O_(2) to 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

`2 N _(2) + 3 O _(2) to 2 N _(2) O _(3),(%` yield = 80%)
Initial mole of `N _(2) = ( 7. 0)/( 28) = 0. 25` mol Moles of `N _(2)` converted `= 0. 25 xx (80)/( 100) = 0.2`
`2 mol I _(2) = 3 mol O _(2) 0.2 mol N _(2) = 0. 3 mol O_(2) (1 mol O _(2) = 2` oxygen atoms)
`=2 xx 0.3 ` mol O atom `= 2 xx 0.3 xx 6. 02 xx 10 ^(23) = 3. 6 xx10 ^(23)`