# How many grams of NH4NO3 are required to produce 33.0 g of N20

Laughing gas (nitrous oxide,N20) is sometimes used as an anesthetic in dentistry. It is produced when ammonium nitrate is decomposed according to the following equation. NH4NO3-> N20 +2H20

How many grams of NH4NO3 are required to produce 33.0 g of N20?

How many grams of water are produced in this reaction?

Get help immediately by chatting with an live tutor right now

**If you find this answer useful please share it with other students.**

33.0g N

_{2}O x (1 mol N_{2}O / 44g N_{2}O) x (1 mol NH_{4}NO_{3}/ 1 mol N_{2}O) x (80g NH_{4}NO_{3}/ 1 molNH_{4}NO_{3}) = Mass of NH_{4}NO_{3}Try to solve for the mass of water produced using the same logic.

chrisf

Tue, 2009-01-27 20:09

Here is an answer to the first part of your question:

1. You need 33 grams of N20, so convert this into moles of N20 using the molar mass

(33 g N2O)(1 mol N2O/30 g N2O) = 1.1 mol N20

2. Since the ratio of NH4NO3 to N2O is 1:1, and you need 1.1 moles of N20, then this means you need 1.1 moles of NH4NO3, and then you can convert this value into grams using the molar mass of NH4NO3

(1.1 mol NH4NO3)(66 g NH4NO3/1 mol NH4NO3) = 72.6 g NH4NO3

TheJoker

Tue, 2009-01-27 20:16

Sorry i used the wrong molar mass for N2O,,, the guy above me is right

TheJoker

Tue, 2009-01-27 20:25

How many grams of water are produced in this reaction?

CARMELCORN1

Tue, 2009-01-27 20:27

Sorry about part one just do everything I did but then use 44 g/mol for molar mass of N20 instead of 30 like i did.

To find the grams of water produced:

you want 33 g of N2O so this means you want

(33 g N2O)(1 mol N2O/44 g N2O) = .75 mol N20

then just using a ratio from the balanced equation:

(.75 mol N2O / x mol H20) = (1 mol N2O / 2 mol H20)

solving for x gives you 1.5 moles of H20, then you can find the grams of water using the molar mass:

1.5 mol H20 (18 g H20/1 mol H20) = 27 g H20

im pretty sure this is right haha

TheJoker

Tue, 2009-01-27 20:56

If it were me, I'd do this:

Mass of NH

_{4}NO_{3}x (1 mol NH_{4}NO_{3}/ 80g NH_{4}NO_{3}) x (1 mol H_{2}O / 1 mol NH_{4}NO_{3}) x (18.02g H_{2}O / 1 mol H_{2}O) = Mass of H_{2}Ochrisf

Tue, 2009-01-27 21:46

that works too

TheJoker

Tue, 2009-01-27 22:24