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Chapter 8: Problem 108

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) \(2.5 \%\)

Short Answer

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Answer: 50%

Step by step solution

01

Find the number of moles of CO₂ formed

We are given the volume of CO₂ formed (1120 cm³) at 0°C and 1 atm. To find the moles of CO₂, we'll use the ideal gas equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. At 0°C, the temperature in Kelvin would be 273 K. The gas constant R is generally given as 0.0821 L atm / K mol. Convert the volume of CO₂ to liters: 1120 cm³ = 1.12 L Now we can solve for the number of moles (n) as follows: PV = nRT ⇨ n = PV / RT n = (1 atm)(1.12 L) / (0.0821 L atm / K mol)(273 K) = 0.05 mol
02

Calculate the moles of CaCO₃

The balanced chemical equation for the reaction between CaCO₃ and HCl is: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) From the stoichiometry, we can see that 1 mole of CaCO₃ reacts to form 1 mole of CO₂. Since we found 0.05 mol of CO₂, we must have had 0.05 mol of CaCO₃ in the marble.
03

Calculate the mass and percentage of CaCO₃

With the moles of CaCO₃ calculated, we can now find the mass. The molar mass of CaCO₃ is: (40.08 g/mol for Ca) + (12.01 g/mol for C) + (3 × 16 g/mol for O) = 100.09 g/mol Now, multiply the moles of CaCO₃ by its molar mass: Mass of CaCO₃ = (0.05 mol)(100.09 g/mol) = 5 g Finally, find the percentage of CaCO₃ in the marble: Percentage of CaCO₃ = (mass of CaCO₃ / total mass of marble) × 100 Percentage of CaCO₃ = (5 g / 10 g) × 100 = 50% Based on our calculations, the percentage of CaCO₃ in the marble is 50%, so the correct answer is (c) 50%.

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Most popular questions from this chapter

The concentration of same aqueous solution of glucose is determined by two students-Sawan and Gautam. Sawan reported the concentration as \(20 \%\) (w/w) and Gautam reported the concentration as \(25 \%(\mathrm{w} / \mathrm{v}) .\) If both the concentrations are correct, then the density of solution is (a) \(0.8 \mathrm{~g} / \mathrm{ml}\) (b) \(1.0 \mathrm{~g} / \mathrm{ml}\) (c) \(1.25 \mathrm{~g} / \mathrm{ml}\) (d) \(1.33 \mathrm{~g} / \mathrm{m} 1\)

It was found from the chemical analysis of a gas that it has two hydrogen atoms for each carbon atom. At \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), its density is \(1.25 \mathrm{~g}\) per litre. The formula of the gas would be (a) \(\mathrm{CH}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{8}\)

Most abundant element dissolved in sea water is chlorine at a concentration of \(19 \mathrm{~g} / \mathrm{kg}\) of sea water. The volume of earth's ocean is \(1.4 \times 10^{21} 1\). How many g-atoms of chlorine are potentially available from the oceans? Density of sea water is \(1 \mathrm{~g} / \mathrm{ml} .\left(N_{\mathrm{A}}=6 \times 10^{23}\right)\) (a) \(7.5 \times 10^{20}\) (b) \(27 \times 10^{21}\) (c) \(27 \times 10^{24}\) (d) \(7.5 \times 10^{19}\)

A quantity of \(4.35 \mathrm{~g}\) of a sample of pyrolusite ore, when heated with conc. HCl, gave chlorine. The chlorine, when passed through potassium iodide solution, liberated \(6.35 \mathrm{~g}\) of iodine. The percentage of pure \(\mathrm{MnO}_{2}\) in the pyrolusite ore is \((\mathrm{Mn}=55, \mathrm{I}=127)\) (a) 40 (b) 50 (c) 60 (d) 70

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 molality of solution 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
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