# In the reaction N_2 + 3O_2 -> 2NO_3, how many grams of N_2 are required to produce 10 grams of NO_3?

## How many moles of $N {O}_{3}$ can be produced from 8 grams of ${O}_{2}$? How many grams of ${O}_{2}$ are required to react with .5 moles of ${N}_{2}$?

##### 1 Answer
Apr 11, 2016

This question is related to Stoichiometry. Lets us first calculate the molar ratio of ${N}_{2}$ and N${O}_{3}$.

As per the equation , one mole of ${N}_{2}$ produces on reaction with 3 moles of ${O}_{2}$ produces 2 moles of N${O}_{3}$.

2 moles of Oxygen = 64 g of Oxygen, is required to react with One mole of Nitrogen or 28 g of Nitrogen and produces 3 moles of N${O}_{3}$ or 62 g of N${O}_{3}$.

so when we use 8 g of Oxygen , we will get ,

Mass of N${O}_{3}$ = ( 62 g of N${O}_{3}$ / 64 g of ${O}_{2}$ ) x 8 g of ${O}_{2}$ = 7. 75 g of N${O}_{3}$.

To react with One mole of ${N}_{2}$ , one needs 3 moles of ${O}_{2}$ , if we begin with half a mole of ${N}_{2}$ , we will need 1.5 moles of ${O}_{2}$.

28 g of ${N}_{2}$ produces 62 g of N${O}_{3}$, in order to get 10 g of N${O}_{3}$ , we will need

28 g of ${N}_{2}$ x 10 g of N${O}_{3}$ / 62 g of N${O}_{3}$

= 4.50 g of ${N}_{2}$.