# What mass of oxygen is required to make 90*g of NO?

Jan 26, 2017

Approx. $42 \cdot g$

#### Explanation:

We could assume either atoms or molecules; for this problem it is wise to use atomic mass.

$\text{Atoms of oxygen"-=(48*g)/(16.00*g*mol^-1)=3.0*mol*"oxygen atoms, i.e "3xxN_A" oxygen atoms.}$

So we need $3.0 \cdot m o l$ $\text{nitrogen atoms}$, or $3 \times {N}_{A}$ $\text{nitrogen atoms,}$ which has an equivalent mass of,

3*molxx14.01*g*mol^-1=??*g