In: Chemistry
A) According to the (unbalanced) reaction: NH3(g) + O2(g) ----> NO2 (g) + H20 (l)
What volume of NO2(g) will be produced from the combustion of 521 g of O2(g), assuming the reaction takes place at standard temperature and pressure?
B) What mass of water will be produced if a 38.1 L sample of ammonia at 1.13 atm of pressure and. 23.1 degree celcius reacts completely?
A)
Molar mass of O2 = 32 g/mol
mass of O2 = 5.21*10^2 g
mol of O2 = (mass)/(molar mass)
= 5.21*10^2/32
= 16.28 mol
Balanced chemical equation is:
4NH3 + 7O2 = 4NO2 + 6H2O
According to balanced equation
mol of NO2 formed = (4/7)* moles of O2
= (4/7)*16.28
= 9.304 mol
At STP, molar volume is 22.4 L/mol
Now use:
volume = number of mol * molar volume
= 9.304 mol * 22.4 L/mol
= 208 L
Answer: 208 L
B)
Given:
P = 1.13 atm
V = 38.1 L
T = 23.1 oC
= (23.1+273) K
= 296.1 K
find number of moles using:
P * V = n*R*T
1.13 atm * 38.1 L = n * 0.08206 atm.L/mol.K * 296.1 K
n = 1.772 mol
From equation,
mol of H2O = (6/4)*number of mol of NH3 reacted
= (6/4)* 1.772
= 2.658 mol
Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol
use:
mass of H2O,
m = number of mol * molar mass
= 2.658 mol * 18.02 g/mol
= 47.89 g
Answer: 47.9 g