If 27g of water is formed during complete combusiton of pure propene `(C_(3)H_(6))`, the mass of propene burnt is

Options

A. 42g
B. 21g
C. 14g
D. 56g

Answer

B

Explanation

Required reaction :
`underset("(Propene)")(C_(3)H_(6))+(9)/(2)O_(2) rarr 3CO_(2) + underset((3 xx 18 = 54))(3H_(2)O)`
Given, `H_(2)O` (formed) = 27g
To find the mass of propene
The mass ratio between `C_(3)H_(6)` and `H_(2)O`
=42: 54
`because` 54 g of `H_(2)D` require `= 42g C_(3)H_(6)`
`therefore 27 g` of `H_(2)O` require
`= (42 xx 27)/(54) = 21g.C_(3)H_(6)`