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Chapter 8: Problem 114

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) \(0.75\)

Short Answer

Expert verified
Answer: The number of moles of carbon in the mixture is 0.5.

Step by step solution

01

Write down the chemical reaction for complete combustion of carbon and sulphur

Combustion of carbon and sulphur produces CO2 and SO2, respectively. The involved chemical reactions are: For carbon: C + O2 -> CO2 For sulphur: S + O2 -> SO2
02

Express the moles of each substance

Let's assume the mixture consists of x g of carbon (C) and (14 - x) g of sulphur (S). Now, using the molar masses of C (12 g/mol) and S (32 g/mol), we can express each substance in terms of moles: Moles of carbon (C) = \(\frac{x}{12}\) Moles of sulphur (S) = \(\frac{14-x}{32}\)
03

Apply the given volume ratio to the moles of \(\mathrm{CO}_{2}\) and \(\mathrm{SO}_{2}\)

We know that the volume ratio of CO2 to SO2 is 2:1. According to Avogadro's law, equal volumes of gases under the same conditions of temperature and pressure contain equal number of molecules or moles. Therefore, the moles ratio is also 2:1. So: \(\frac{\text{moles of CO2}}{\text{moles of SO2}}\) = \(\frac{2}{1}\) Since all carbon in the mixture is converted to CO2 and all sulphur is converted to SO2, we can use the relationship between moles of C and S from Step 2: \(\frac{\frac{x}{12}}{\frac{14-x}{32}}\) = \(\frac{2}{1}\)
04

Solve the equation for x

From the equation in Step 3, we have: \(\frac{x}{12}\) = \(\frac{2(14-x)}{32}\) To solve for x: 32x = 24(14-x) 32x = 336 - 24x 56x = 336 x = 6
05

Calculate the moles of carbon in the mixture

Since x = 6, we can now find the moles of carbon (C) in the mixture: Moles of carbon (C) = \(\frac{6}{12}\) = 0.5 The moles of carbon in the mixture is 0.5, which corresponds to the answer (c).

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Most popular questions from this chapter

A quantity of \(10 \mathrm{~g}\) of a piece of marble was put into excess of dilute \(\mathrm{HCl}\)acid. When the reaction was complete, \(1120 \mathrm{~cm}^{3}\) of \(\mathrm{CO}_{2}\) was obtained at \(0^{\circ} \mathrm{C}\) and 1 atm. The percentage of \(\mathrm{CaCO}_{3}\), in the marble is (a) \(5 \%\) (b) \(25 \%\) (c) \(50 \%\) (d) \(2.5 \%\)

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