Question

17. What would be percentage composition by volume of a mixture of CO and CH, whose 10.5 mL requires
9 mL oxygen for complete combustion?
(1) CO = 80% CH, = 20%
(2) CO = 90% CH4 = 10%
(3) CO = 76.2% CH, = 23.8%
(4) CO = 66% CH. = 34%​

Answer 1

Answer:

(2).CO=90%,CH4=10% .

Hope it helps you.

Answer 2

Answer:

(1) CO = 80% CH, = 20% (2) CO = 90% CH4 = 10%