The combustion of carbon(II)oxide in oxygen can be represented by equation:
2CO + O\(_2\) → 2CO\(_2\)
Calculate the volume of the resulting mixture at the end of the reaction, if 50cm\(^3\) of carbon(ii)oxide was exploded in 100cm\(^3\) of oxygen
75cm\(^3\)
125cm\(^3\)
100cm\(^3\)
50cm\(^3\)
Explanation
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2CO + O2 ~ 2CO2
2 : 1 : 2
50cm^3 : 100cm^3: 50cm^3
50cm^3 : 25cm^3 : 25cm^
Vol of resulting mixture 100+25 =125cm^3 ans B
when they ask for volume of resulting mixture they imply the total volume of the product plus unreacted gas volume
which makes up thr outcome of the reaction.
2CO + O2 --> 2CO2
50cm3 : 25cm3 : 50cm3
therefore, unreacted O2 volume = 100cm3 - 25cm3 = 75cm3
and product volume = 50cm3.
....resulting mixture volume = 75cm3 + 50cm3 = 125cm3 
2CO2. + O2 - 2CO2
before reaction: 50cm^3 : 100cm^3 : 0cm^3
during reaction: 50cm^3: 25cm^3 : 50cm^3
end of reaction: - : 75cm^3: 50cm^3
residual gas=volume of
unreacted + carbon(IV)oxide formed
oxygen
75cm^3 + 50cm^3 =125cm^3
if 2moles of Co is 50, then 1mole of Co will be 25.
if 1mole of O2 is 100, then 2mole will be 200.
the simple way to understand is by Minusing the bigger value from the smaller value.
2mole co --50, 1mole Co 25.
2mole O2--200, 1mole O2 100.
from the above CO have the smaller values. therefore we are using the values for Co to minus through.
2Co. +O2 ----> 2CO2
50. 100. ----
50. 25. 50
—. —. —
0 75. 50 … therefore the value left is 75 and 50, add up=125.
Note: where ever you see 2mole, you use the value 50. where ever you see 1mole, you use the value 25. this is because we are using the values of Co to minus.
