find the volume of NO2 and O2 gases produced at STP when a sample of 54g of N2O5 is heated
Answers
Dear Student,
The balance equation for the decomposition of N2O5 can be written as:
2 N2O5(g) → 4 NO2(g) + O2(g)
Given:
Mass of O2(g) formed = 1.60 g
Molar mass of O2 = 16.0 g/mol
Moles of O2 = Mass / Molar mass = 1.60 / 16.0 = 0.1 moles
The mole ratio of O2 and NO2 is 1 : 4 hence moles of NO2 = 0.1 x 4 = 0.4 moles of NO2
Molar mass of NO2 = 46.00 g/mol
Mass of NO2 = 46.00 x 0.4 = 18.4 g
Answer: 18.4 g of NO2 will form..
Explanation:
thanku..
Mark as brainliest
The thermal decomposition of N2O5 follows the given equation,
2N2O5 -----------> 4NO2 + O2
molar mass of N2O5 = 2*14 + 5*16 =>28+80 => 108 grams per mole.
So, 54 grams of N2O5 contains 54/108 => 0.5 moles of N2O5
From the equation it is visible that, 2 parts of N2O5 produces 4 parts of NO2 and 1 part of O2.
So, 2 parts of N2O5=0.5 moles
So, 1 part= 0.5/2 => 0.25 moles.
1 mole of any gas at STP has a volume of 22.4 litres.
Hence, 4 parts of NO2 will be 4 X 0.25 => 1 mole of NO2 gas.
1 X 22.4 litres => 22.4 litres of reddish-brown NO2 gas is released.
Hence, 1 part of O2 will be 1 X 0.25 => 0.25 moles of O2 gas.
0.25 X 22.4 litres => 5.6 litres of colourless oxygen gas is released.