Question

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 `NO_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`?

Answer 1

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`.