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)`.
3.24×10^23
3.6×10^23
1.8×10^23
6.02×10^23

Options

A. `3.24xx10^(23)`
B. `3.6xx10^(23)`
C. `1.8xx10^(23)`
D. `6.02xx10^(23)`

Answer

B

Explanation

`2N_(2)+3O_(2) to 2N_(2)O_(3)`, (%yeild =80%)
Initial mol of `N_(2)=(7.0)/(28)=0.25 mol" "" Moles of "N_(2)` converted `=0.25xx(80)/(100)=0.2`
2 mol `N_(2)=3 mol O_(2)" "0.2 mol N_(2)=0.3 mol O_(2) (1mol O_(2)=2` oxygen atoms)
`=2xx0.3" mol O atom "=2xx0.3xx6.02xx10^(23) =3.6xx10^(23).`