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Chapter 8: Problem 1

A quantity of \(10 \mathrm{~g}\) of a hydrocarbon exactly requires \(40 \mathrm{~g}\) oxygen for complete combustion. The products formed are \(\mathrm{CO}_{2}\) and water. When \(\mathrm{CO}_{2}\) gas formed is absorbed completely in lime water, the mass of solution increases by \(27.5 \mathrm{~g}\). What is the mass of water formed in combustion? (a) \(22.5 \mathrm{~g}\) (b) \(27.5 \mathrm{~g}\) (c) \(50 \mathrm{~g}\) (d) \(10 \mathrm{~g}\)

Short Answer

Expert verified
Answer: The mass of water formed in the complete combustion of the hydrocarbon is \(22.5 \mathrm{~g}\).

Step by step solution

01

Write the balanced chemical equation

For any hydrocarbon (represented by general formula CxHy), the complete combustion reaction can be represented as: CxHy + O2 -> n*CO2 + m*H2O Here, n and m represent the stoichiometric coefficients that we don't yet know. Balancing the equation is not important in this particular problem since it doesn't involve finding the unknown as part of the exercise.
02

Determine the mass of CO2 absorbed in lime water

We know that the mass of the lime water solution increases by 27.5 g when CO2 is absorbed. This means the mass of CO2 formed in the reaction is also 27.5 g.
03

Calculate the total mass of products formed during combustion

According to the Law of Conservation of Mass, the mass of all reactants together must be equal to the mass of all products together. So, in this case, Mass of reactants = Mass of hydrocarbon + Mass of oxygen = 10 g + 40 g = 50 g Therefore, the total mass of products (CO2 and H2O) formed during the combustion = 50 g
04

Determine the mass of water formed in the reaction

We now know the mass of CO2 and the total mass of products. We can find the mass of water formed by subtracting the mass of CO2 from the total mass of products. Mass of water formed = Mass of products - Mass of CO2 = 50 g - 27.5 g = 22.5 g So the mass of water formed in the combustion is 22.5 g. Hence, the correct answer is (a) \(22.5 \mathrm{~g}\).

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Most popular questions from this chapter

A \(1.50 \mathrm{~g}\) sample of potassium bicarbonate having \(80 \%\) purity is strongly heated. Assuming the impurity to be thermally stable, the loss in weight of the sample, on heating, is (a) \(3.72 \mathrm{~g}\) (b) \(0.72 \mathrm{~g}\) (c) \(0.372 \mathrm{~g}\) (d) \(0.186 \mathrm{~g}\)

One mole of a mixture of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) requires exactly \(20 \mathrm{~g}\) of \(\mathrm{NaOH}\) in solution for complete conversion of all the \(\mathrm{CO}_{2}\) into \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) How many grams more of \(\mathrm{NaOH}\) would it require for conversion into \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) if the mixture (one mole) is completely oxidized to \(\mathrm{CO}_{2} ?\) (a) \(60 \mathrm{~g}\) (b) \(80 \mathrm{~g}\) (c) \(40 \mathrm{~g}\) (d) \(20 \mathrm{~g}\)

When burnt in air, \(14.0 \mathrm{~g}\) mixture of carbon and sulphur gives a mixture of \(\mathrm{CO}_{2}\) and \(\mathrm{SO}_{2}\) in the volume ratio of \(2: 1\), volume being measured at the same conditions of temperature and pressure. Moles of carbon in the mixture is (a) \(0.25\) (b) \(0.40\) (c) \(0.5\) (d) \(0.75\)

Two successive reactions, \(\mathrm{A} \rightarrow \mathrm{B}\) and \(\mathrm{B} \rightarrow \mathrm{C}\), have yields of \(90 \%\) and \(80 \%\) respectively. What is the overall percentage yield for conversion of \(\mathrm{A}\) to \(\mathrm{C}\) ? (a) \(90 \%\) (b) \(80 \%\) (c) \(72 \%\) (d) \(85 \%\)

When \(500 \mathrm{ml} \mathrm{CO}_{2}\) gas is passed through red hot charcoal, the volume becomes \(700 \mathrm{ml}\). The volume of \(\mathrm{CO}_{2}\) converted into \(\mathrm{CO}\) is (a) \(200 \mathrm{ml}\) (b) \(300 \mathrm{ml}\) (c) \(350 \mathrm{ml}\) (d) \(500 \mathrm{ml}\)
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