NH₄NO₃ → N₂O + 2H₂O When 45.70 g of NH₄NO₃ decomposes, what mass of each product is formed?

25.13 g of  and 20.56 g of will be produced from 45.70 g of

Explanation:

To calculate the moles :

The balanced chemical equation is:

According to stoichiometry :

1 mole of produce = 1 mole of

Thus 0.571 moles of will require=  of  

Mass of

1 mole of produce = 2 moles of

Thus 0.571 moles of will require=  of  

Mass of

Thus 25.13 g of  and 20.56 g of will be produced from 45.70 g of

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

yes it is balanced