Chemistry, asked by PickleFlicker6789, 1 year ago

12 ml of a gaseous hydrocarbon was exploded with 50 ml of oxygen the volume after explosion bus 32 mm on treatment with cable has the volume reduced to 1 ml then the formula of hydrocarbon

Answers

Answered by nuclearscientist321
3

Silver articles become black when exposed to air

Answered by itemderby
2

Explanation:

Chemical equation that represents combustion of a hydrocarbon is as follows.

    C_{x}H_{y} + O_{2} \rightarrow CO_{2} + H_{2}O

It is given that volume of carbon dioxide and unreacted oxygen is equal to 32 ml.

As carbon dioxide gas is absorbed by potassium hydroxide. So, when 32 ml of this mixture is treated with KOH then it is reduced to 1 ml.

Hence, amount of CO_{2} formed is (32 ml - 1 ml) = 33 ml.

Therefore, amount of unreacted oxygen = 1 ml

Volume of reacted oxygen = 50 ml - 1 ml = 49 ml

Volume of hydrocarbon is given as 12 ml.

Therefore, applying the principle of atom conservation on C as follows.

  moles of C_{x}H_{y} = 1

No. of moles of CO_{2} will be calculated as follows.

             x \times 12 ml = 1 \times 33 ml

                    x = 2.7 = 3 (approx)

Now, applying the principle of atom conservation on H as follows.

    y \times {\text{no. of moles of}} C_{x}H_{y} = 2 \times {\text{no. of moles of}} H_{2}O   ....... (1)

Now, applying the principle of atom conservation on O as follows.

      no. of moles of O_{2} = 1 \times /text{no. of moles of} H_{2}O + 2 \times /text{no. of moles of} CO_{2}[/tex]

No. of moles of O_{2} = \frac{y}{2} \times no. of moles of C_{x}H_{y} + 2 \times no. of moles of CO_{2}

            2 \times 49 ml = \frac{y}{2} \times 12 + 2 \times 33 ml

                      y = 6

Hence, molecular formula of the given hydrocarbon is C_{3}H_{6}.

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