What is the volume of nitrogen that will be produced at stp from the decomposition of 9.60g of ammonium dioxonitrate(III)?

Answer 1

how many moles of NH4NO2 is 9.60g?
you'll get that many moles of N2
each mole occupies 22.4L at stp

Answer 2

I. Ammonium dioxonitrate(III) solution - NH4NO2 readily decomposes slowly on slight warming at ordinary temperature to liberate nitrogen.
NH4NO2----N2+2H2O

Molar mass of NH4NO2=(14+4+14+32)=64g

64g of NH4NO2 at s.t.p produces 22.4dm^3
so therefore,
9.6g of NH4NO2 at s.t.p will produce = (9.6*22.4)/64=3.36dm^3 ( answer)

Answer 3

II. To calculate the number of mole of NH4NO2 of 9.6g, we first calculate the "molar mass" of the compound

Key Points

1. The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance (mol).

OR

2. The molar mass of a compound can be calculated by adding the standard atomic masses (in g/mol) of the constituent atoms.

Since we are to find the number of mole of NH4NO2, we go with number 2

where;
Mass number of Nitrogen (N)=14g
Mass number of Hydrogen (H)= 1g
Mass number of Oxygen (O)=16g

so therefore,
Molar mass of NH4NO2=14+(1*4)+14+(16*2)=64g

NOTE: 1 mole of a substance=22.4dm^3 or 22.4L

So therefore,
NH4NO2 contains 1 mole which is = 22.4L

Finally,
64g of NH4NO2 contains 1 mole
Therefore, 9.60g of NH4NO2 contains = (9.60*1)/64=0.15 mole (answer)

Answer 4

oops vek Tor--
Obviously you didn't mean this.
"So therefore,
NH4NO2 contains 1 mole which is = 22.4L"
NH4NO2 is a solid and the bit about 22.4L/mol is for a gas @ STP.