Question

1g of a gaseous hydrocarbon CxHy on
combustion consumes 3.636g of oxygen
and gave 3g of CO2. What is the value of y?
(Mol. Wt. of CxHy is 44)​

Answer 1

The value of y is 26

Explanation:

Given:

1 g of gaseous hydrocarbon $C_xH_y$ on combustion consumes 3.636 g of Oxygen and gives 3 g of CO₂

Molecular weight of $C_xH_y$ is 44

To find out:

The value of y

Solution:

Molecular weight of O₂ = 32

Molecular weight of CO₂ = 44

Given that Molecular weight of $C_xH_y$ is 44

Hence

$6x+y=44$

$\implies y=44-6x$   .  . . . . . . . . . (1)

Moles of O₂ = Weight/Molecular weight = 3.636/32 = 0.113

Moles of CO₂ = 3/44

Moles of $C_xH_y$  = 1/44

The balanced equation will be

$C_xH_y+(x+\frac{y}{2})O_2\rightarrow xCO_2+yH_2O$

1 moles of $C_xH_y$ gives x moles of CO₂

1/44 moles of $C_xH_y$ gives x/44 moles of CO₂

Thus,

x/44 = 3/44

or, x = 3

Therefore, from equation (1)

$y=44-6\times 3$

$y=44-18$

$\implies y=26$

Hope this answer is helpful.

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