How much NH4NO3 is required to produce 47.2 g of N2O? (more detail plz read on)

Laughing gas (nitrous oxide, N2O) is sometimes used as an anesthetic in dentistry. It is produced when ammonium nitrate is decomposed according to the reaction:
NH4NO3(s) -----> N2O(g) + 2H2O(l)
How much NH4NO3 is required to priduce 47.2 g of N2O?
Answer in units of g thank you!

NH4NO3 (s) -----> N2O (g) + 2H2O (l)

moles N2O = mass / molecular weight = 47.2 g / 44.0 g/mol = 1.07 moles of N2O

1.07 moles N2O x (1 mole NH4NO3 / 1 mole N2O) = 1.07 moles of NH4NO3

1.07 moles of NH4NO3 x 80.0 g/mol = 85.6 grams of NH4NO3


Hope this helps...good luck!

47.2 * (14+4+14+3*16)/(2*14+16)= 47.2 *90/44 = 96.6 g