Question

10 g mixture of ethylene and ethane gases can be decolorize 200 g of 16?

Answer 1

The student is asking about the decolorization of a mixture of ethylene and ethane gases. Assuming a 100% yield, the amount of ethylene produced can be calculated using the molar masses of ethane and ethylene.

Step-by-step explanation:

In this question, the student is asking about the decolorization of a mixture of ethylene and ethane gases. It seems that there might be a typo in the question, as '16' does not fit in the context of the given information. Nonetheless, the pyrolysis reaction of ethane produces ethylene and hydrogen gas. If we assume a 100% yield, then the number of grams of ethylene produced can be calculated by using the molar masses of ethane and ethylene:

Molar mass of ethane (C2H6) = 30.07 g/mol

Molar mass of ethylene (C2H4) = 28.05 g/mol

Given that 10 g of the mixture contains ethane and ethylene, we can set up the following equation:

10 g * (28.05 g/mol / 30.07 g/mol) = x g of ethylene

By performing the calculations, we can find the amount of ethylene produced in the mixture.

Answer 2

When 1.000 × 10³ kg of ethane is pyrolyzed, assuming a 100.0% yield, it will produce the same amount of ethylene (1.000 × 10³ kg).

Step-by-step explanation:

The pyrolysis of ethane produces ethylene and hydrogen gas.

The balanced chemical equation for the reaction is:
C₂H₆ → C₂H₄ + H₂

So, when 1.000 × 10³ kg of ethane is pyrolyzed, assuming a 100.0% yield, it will produce the same amount of ethylene (1.000 × 10³ kg).