Question

25g of an unknown hydrocarbon upon burning produces 88g of \[C{O_2}\] and 9g of \[{H_2}O\]. This unknown hydrocarbon contains.
A. 20g of carbon and 5g of hydrogen
B. 24g of carbon and 1g of hydrogen
C. 18g of carbon and 7g of hydrogen
D. 22g of carbon and 3g of hydrogen

Answer 1

Hint- We need to calculate the amount of carbon and hydrogen in the product and then use that information to calculate the amount of carbon and hydrogen in 25g of unknown hydrocarbon which is burned.

Complete step by step answer:
Let’s start the answer by calculating the amount of carbon and hydrogen in the product that is being formed.
So, 88g of \[C{O_2}\] and 9g of \[{H_2}O\] is being formed.
The molecular mass of CO2 is 44g consisting of 12g of Carbon (C) and 32g of Oxygen (O) since one mole of O weighs 16g. So, the product is having 2 moles of \[C{O_2}\] which means that 24g of Carbon is present in the product.
Similarly, in the case of water \[{H_2}O\], the molecular weight of \[{H_2}O\] is 18g comprising 16g of oxygen (O) and 2g of Hydrogen (H). The product is having half a mole of \[{H_2}O\], which means that only 1g of hydrogen is present in the product.
One should know that for the burning process only oxygen is required and not hydrogen or carbon dioxide, which means that these two gases should not have interfered with the composition of the final product.
We also know that the amount of atoms present in the product is similar to that of reactants. So, the unknown hydrocarbon is having 24g of carbon and 1g of hydrogen as only this much amount of carbon and hydrogen is being present in the product.
The answer to this question is B. 24g of carbon and 1g of hydrogen.

Note: We must know that the \[C{O_2}\] is always produced in hydrocarbon combustion reactions. Producing \[C{O_2}\] and \[{H_2}O\] is actually how useful energy is obtained from fossil fuels. This is the main reason we need to distinguish between carbon dioxide and other waste products that obtain from impurities in the fuel such as sulphur(S) and nitrogen (N) compounds.
While answering such questions the student should know the molecular weight of the final products as well as the individual weight of the atoms which are present in the molecule. This entire area is required for better understanding and fast answering of questions.