# 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?

### Re: homework problems on Ideal Stoichiometric Calculations

33.0g N2O x (1 mol N2O / 44g N2O) x (1 mol NH4NO3 / 1 mol N2O) x (80g NH4NO3 / 1 molNH4NO3) = Mass of NH4NO3

Try to solve for the mass of water produced using the same logic.

### Re: homework problems on Ideal Stoichiometric Calculations

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

### Re: homework problems on Ideal Stoichiometric Calculations

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

### Re: homework problems on Ideal Stoichiometric Calculations

How many grams of water are produced in this reaction?

### Re: homework problems on Ideal Stoichiometric Calculations

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

### Re: homework problems on Ideal Stoichiometric Calculations

If it were me, I'd do this:

Mass of NH4NO3 x (1 mol NH4NO3 / 80g NH4NO3) x (1 mol H2O / 1 mol NH4NO3) x (18.02g H2O / 1 mol H2O) = Mass of H2O

that works too