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Chapter 8: Problem 49

Equal masses of oxygen, hydrogen and methane are taken in identical conditions. What is the ratio of the volumes of the gases under identical conditions? (a) \(16: 1: 8\) (b) \(1: 16: 2\) (c) \(1: 16: 8\) (d) \(2: 16: 1\)

Short Answer

Expert verified
A: The volume ratio is 1:16:2.

Step by step solution

01

Determination of molar mass of the gases

For determining the molar mass of each gas, we sum up the atomic masses of the elements in the gas: 1. Oxygen (O\(_2\)) - The atomic mass of oxygen is 16. Since oxygen is diatomic, the molar mass of the O\(_2\) molecule will be \(2 \times 16 = 32\) g/mol. 2. Hydrogen (H\(_2\)) - The atomic mass of hydrogen is 1. Being diatomic, the molar mass of an H\(_2\) molecule is \(2 \times 1 = 2\) g/mol. 3. Methane (CH\(_4\)) - The atomic mass of carbon is 12, and hydrogen is 1. The molar mass of a CH\(_4\) molecule will be \(12 + 4 \times 1 = 16\) g/mol.
02

Application of Avogadro's Law

According to Avogadro's law, equal volumes of gases at the same temperature and pressure contain the same number of molecules. Therefore: \(\text{Volume} \propto \frac{\text{mass}}{\text{molar mass}}\) Now, we are given equal masses of all three gases, so we can write the ratio of the volumes as: \(\text{Volume ratio} = \frac{\text{mass of O}_2}{\text{molar mass of O}_2} : \frac{\text{mass of H}_2}{\text{molar mass of H}_2} : \frac{\text{mass of CH}_4}{\text{molar mass of CH}_4}\) Since the masses of gases are the same, we can write the volume ratio as: \(\text{Volume ratio} = \frac{1}{\text{molar mass of O}_2} : \frac{1}{\text{molar mass of H}_2} : \frac{1}{\text{molar mass of CH}_4}\)
03

Calculate the volume ratio

Using the molar masses from Step 1, plug in the values into the formula: \(\text{Volume ratio} = \frac{1}{32} : \frac{1}{2} : \frac{1}{16}\) To find the smallest whole number ratio, multiply the fractions by their least common multiple, which is 32: \(\text{Volume ratio} = \frac{1 \cdot 32}{32} : \frac{32}{2} : \frac{2 \cdot 32}{16} = 1:16:2\) Hence, the correct answer is (b) \(1:16:2\).

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Most popular questions from this chapter

Air contains \(20 \% \mathrm{O}_{2}\), by volume. What volume of air is needed at \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) for complete combustion of \(80 \mathrm{~g}\) methane? (a) 101 (b) 501 (c) 2241 (d) 11201

A quantity of \(10 \mathrm{~g}\) of acetic acid is dissolved in \(100 \mathrm{~g}\) of each of the following solvents. In which solvent, the mole fraction of solute is maximum? Assume no any dissociation or association of acetic acid in the solvent. (a) Water (b) Ethanol (c) Benzene (d) Same in all solvents

Hydrogen cyanide, HCN, is prepared from ammonia, air and natural gas \(\left(\mathrm{CH}_{4}\right)\) by the following process. \(2 \mathrm{NH}_{3}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{CH}_{4}(\mathrm{~g}) \stackrel{\mathrm{Pt}}{\longrightarrow}\) \(2 \mathrm{HCN}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) If a reaction vessel contains \(11.5 \mathrm{~g} \mathrm{NH}_{3}\), \(10.0 \mathrm{~g} \mathrm{O}_{2}\), and \(10.5 \mathrm{~g} \mathrm{CH}_{4}\), what is the maximum mass, in grams, of hydrogen cyanide that could be made, assuming the reaction goes to completion? (a) \(18.26 \mathrm{~g}\) (b) \(5.625 \mathrm{~g}\) (c) \(17.72 \mathrm{~g}\) (d) \(16.875 \mathrm{~g}\)

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) \(0.186 \mathrm{~g}\)

What volume of \(0.8 \mathrm{M}-\mathrm{AlCl}_{3}\) solution should be mixed with \(50 \mathrm{ml}\) of \(0.2 \mathrm{M}-\mathrm{CaCl}_{2}\) solution to get a solution of chloride ion concentration equal to \(0.6 \mathrm{M} ?\) (a) \(5.56 \mathrm{ml}\) (c) \(50 \mathrm{ml}\)
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