Question

40 g of a sample of carbon on combustion left 10% of it
unreacted. The volume of oxygen required at STP for this
combustion reaction is
(a) 22.4 L
(b) 67.2L
(c) 11.2L
(d) 44.8L​

Answer 1

Given:

The mass of the sample of carbon = 40 g

10% of the sample is left unreacted

To find:

The volume of oxygen required at STP for this combustion reaction

Solution:

The combustion reaction that occurs is as follows:

C + O₂ → CO₂

It is given that 10% of 40 g of the sample of carbon is left unreacted

∴ The amount of carbon which reacts from the 40 g sample of carbon during the combustion reaction is,

= 40 - [10% of 40]

= 40 - [$\frac{10}{100} \:\times\:40$]

= 40 - 4

= 36 g

At STP conditions, we know that

The volume of 1 mole of gas is 22.4 L

and

Molecular mass of carbon = 12 g/mol

So, we can say that at STP

12 g of carbon, requires 22.4 L of oxygen for the combustion reaction

∴ The volume of oxygen required by 36 g of the carbon during the combustion, is

= $\frac{22.4}{12}\:\times\:36$

= 22.4 × 3

= 67.2 L ← option (b)

Thus, the volume of oxygen required at STP for this  combustion reaction is 67.2 L.

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Answer 2

correct option is B ..

67.2 L