A mixtures of methane and ethylene in the volume ration x:y has total volume of 30mL. On complete combustion it gave 40mL of `CO_(2)`. IF the ratio of

Question

A mixtures of methane and ethylene in the volume ration x:y has total volume of 30mL. On complete combustion it gave 40mL of `CO_(2)`. IF the ratio of

A mixtures of methane and ethylene in the volume ration x:y has total volume of 30mL. On complete combustion it gave 40mL of `CO_(2)`. IF the ratio of original mixture is y:x instead of x:y. What volume of `CO_(2)` would have been produced a combustion?
A. 50mL
B. 75mL
C. 100mL
D. 125mL

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - A
Combustion of `CH_(4) and C_(2)H_(4)` may be represented as.
`"Combustion of "CH_(4) "may be represented as",`
`underset(xmL)(CH_(4)(g))+2O_(2)(g) to underset(xmL)(CO_(2)(g))+2H_(2)O(g)`
`underset(ymL)(C_(2)H_(4)(g))+3O_(2)(g) to underset(ymL)(2CO_(2)(g))+2H_(2)O(g)`
Given, x+y=30
x+2y=40
`therefore x=20, y=10`
When we take y mL `CH_(4) and "x mL" C_(2)H_(4)`, then volume of `CO_(2)` will be (y+2x), i.e. 50mL.