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#?

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#.