What is the volume of nitrogen that will be produced at stp from the decomposition of 9.60g of ammonium dioxonitrate(III)?
how many moles of NH4NO2 is 9.60g?
you'll get that many moles of N2
each mole occupies 22.4L at stp
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)
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)
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.
To find the volume of nitrogen that will be produced at STP (Standard Temperature and Pressure) from the decomposition of ammonium dioxonitrate(III), we need to follow these steps:
Step 1: Write a balanced chemical equation for the decomposition reaction.
The decomposition of ammonium dioxonitrate(III) can be represented by the following balanced chemical equation:
2 NH4N2O2(s) → 2 N2(g) + 4 H2O(g) + O2(g)
Step 2: Calculate the number of moles of ammonium dioxonitrate(III).
To calculate the number of moles, we use the formula:
moles = mass / molar mass
The molar mass of ammonium dioxonitrate(III) is:
NH4N2O2: 2 x 14.01 g/mol (N) + 4 x 1.01 g/mol (H) + 2 x 16.00 g/mol (O) = 80.06 g/mol
moles = 9.60 g / 80.06 g/mol ≈ 0.120 mol
Step 3: Use stoichiometry to find the number of moles of nitrogen.
From the balanced chemical equation, we see that 2 moles of ammonium dioxonitrate(III) produce 2 moles of nitrogen gas (N2). Therefore, the number of moles of nitrogen is also 0.120 mol.
Step 4: Apply the ideal gas law to find the volume of nitrogen gas.
The ideal gas law states:
PV = nRT
Where:
P = pressure (at STP, it is 1 atm)
V = volume (unknown)
n = number of moles of gas (0.120 mol)
R = gas constant (0.0821 L·atm/(mol·K)) – using atm and mol units
T = temperature (at STP, it is 273 K)
Rearranging the formula to solve for V:
V = (nRT) / P
V = (0.120 mol)(0.0821 L·atm/(mol·K))(273 K) / 1 atm
Calculating:
V = 2.3452 L
Therefore, the volume of nitrogen that will be produced at STP from the decomposition of 9.60 g of ammonium dioxonitrate(III) is approximately 2.35 L.