Q4 When 20 gram of Methane Gas are burnt in presence of 32 gram of oxygen in
a reaction vessel, which compound is limiting reagent?
CH + 2
Answers
Answered by
4
Explanation:
Moles of Methane in 32g=
16
32
=2 moles
Moles of Oxygen in 32g=
32
32
=1 mole
Moles of CO
2
in 20g=
40
20
=0.5 mole
CH
4
+2O
2
⟶CO
2
+2H
2
O
1 mole of Methane react with 2 mole of O
2
1 mole of O
2
will react with
2
1
mole of CH
4
=0.5 mole of CH
4
CH
4
+2O
2
⟶CO
2
+2H
2
O
at t=0 2 1 - -
2−0.5 1−2×0.5 0.5 -
as upon reacting of 0.5 mole of CH
4
; 0.5 mole of CO
2
must be produced.
The produced moles of CO
2
is 0.5
Thus % yield is=
0.5
0.5
×100=100% .
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