Question

how many grams of nitrogen gas will be formed upon the complete reaction of 30.6 grams ammonium nitrate

Answer 1

5.355 g

Step-by-step explanation:

first you have 30.6 g from ammonium nitrate ( NH4NO3 )

molecular weight for NH4NO3 is 80 g/mole

and molecular weight for nitrogen gas N2 is 14 g/ mole

make this

NH4NO3 --------------> N2

80 g/mol --------------> 14 g/mol

30.6 g ---------------> x

So X = 14 x 30.6 ÷ 80 = 5.355 g of N2

Good Luck