Chemistry, asked by shraddhadwivedi72, 7 months ago

30ml CH4 and 90ml O2 is mixed and reacted in eudiometer tube. the resulting gas was passed through KOH solution. find the composition of gas before and after passing through KOH solution

Answers

Answered by Anonymous
4

Answer:

When the mixture is sparked, CO is converted into carbon dioxide.

When this mixture is passed through KOH solution, carbon dioxide reacts to form potassium carbonate.

This corresponds to th decrease in the volume.

Hence, the original composition of the mixture can be either 30 ml, 60ml, 10 ml or 30 ml, 50ml, 20 ml.

In option A, 20 ml of CO may react with 10 ml of oxygen to form 20 ml of carbon dioxide. 60 ml of carbon dioxide are already present. Thus total volume of carbon dioxide will be 80ml.

In option B, 15 ml of CO may react with 30 ml of oxygen to form 30 ml of carbon dioxide. 50 ml of carbon dioxide are already present. Thus total volume of carbon dioxide will be 80ml.

Answered by surya5299
3

Answer:

When the mixture is sparked, CO is converted into carbon dioxide.

When this mixture is passed through KOH solution, carbon dioxide reacts to form potassium carbonate.

This corresponds to th decrease in the volume.

Hence, the original composition of the mixture can be either 30 ml, 60ml, 10 ml or 30 ml, 50ml, 20 ml.

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