Question

What volume of \(\mathrm{CO}_{2}\) measured at STP can be obtained by reacting \(50 \mathrm{~g} \circ \mathrm{f} \mathrm{CaCO}_{3}\) with excess hydrochtoric acid? \(\mathrm{CaCO}_{3}\) ? \(2 \mathrm{IICl} \longrightarrow \mathrm{CaCl}_{2}+\mathrm{II}_{2} \mathrm{O}+\mathrm{CO}_{2}\) (1) \(22.4\) litrcs (2) \(11.2\) litres (3) \(1.12\) litres (4) \(44.8\) litres

Short answer

The volume of \(\mathrm{CO}_{2}\) at STP is 11.2 litres. Option (2).

Step-by-step solution

Write the Balanced Chemical Equation

The balanced chemical equation for the reaction is:\[ \mathrm{CaCO}_{3} + 2\mathrm{HCl} \rightarrow \mathrm{CaCl}_{2} + \mathrm{H}_{2}\mathrm{O} + \mathrm{CO}_{2} \]

Determine the Molar Mass of \(\mathrm{CaCO}_{3}\)

Calculate the molar mass of \(\mathrm{CaCO}_{3}\):\[ \text{Molar mass of } \mathrm{CaCO}_{3} = \text{Mass of } \mathrm{Ca} (40.08) + \text{Mass of } \mathrm{C} (12.01) + \text{Mass of } 3\mathrm{O} (3 \times 16.00) = 100.09 \text{ g/mol} \]

Convert Mass to Moles

Use the molar mass to convert 50 g of \(\mathrm{CaCO}_{3}\) to moles:\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{50 \text{ g}}{100.09 \text{ g/mol}} = 0.5 \text{ mol} \]

Use Stoichiometry to Find Moles of \(\mathrm{CO}_{2}\)

From the balanced chemical equation, 1 mole of \(\mathrm{CaCO}_{3}\) produces 1 mole of \(\mathrm{CO}_{2}\).\[ \text{Moles of } \mathrm{CO}_{2} = 0.5 \text{ mol} \]

Convert Moles of \(\mathrm{CO}_{2}\) to Volume at STP

At STP (Standard Temperature and Pressure), 1 mole of a gas occupies 22.4 litres. So, the volume of \(\mathrm{CO}_{2}\) is:\[ \text{Volume} = 0.5 \text{ mol} \times 22.4 \text{ L/mol} = 11.2 \text{ L} \]

Key concepts

Stoichiometry

Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. It allows us to predict the amounts of substances needed and produced. We start with a balanced chemical equation that shows the exact ratio of reactants to products. In our exercise, the balanced equation is: \[ \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \] Using stoichiometry, we understand that 1 mole of CaCO\(_3\) reacts with 2 moles of HCl to produce 1 mole of CO\(_2\). This ratio is crucial for converting between moles of different substances in the reaction. So, if we know the moles of one substance, we can find the moles of another.

Gas Laws at STP

Standard Temperature and Pressure (STP) refers to a temperature of 0°C (273.15 K) and a pressure of 1 atmosphere (atm). Under these conditions, 1 mole of any ideal gas occupies 22.4 liters. This is known as the molar volume of a gas at STP. In the problem, we use this principle to convert moles of CO\(_2\) into liters. Since we found that 0.5 moles of CaCO\(_3\) produce 0.5 moles of CO\(_2\), we can use the molar volume to find the volume of CO\(_2\) gas produced: \[ \text{Volume} = 0.5 \text{ mol} \times 22.4 \text{ L/mol} = 11.2 \text{ L} \] This relationship helps simplify many gas law calculations, making STP a common reference point in chemistry.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It's crucial for converting between mass and moles—a common step in stoichiometric calculations. To find the molar mass of a compound, you add up the atomic masses of all the atoms in its formula. For calcium carbonate (CaCO\(_3\)):

• Calcium (Ca): 40.08 g/mol
• Carbon (C): 12.01 g/mol
• Oxygen (O): 16.00 g/mol (and there are 3 O atoms)

Hence, the molar mass of CaCO\(_3\) is: \[ 40.08 + 12.01 + (3 \times 16.00) = 100.09 \text{ g/mol} \] This calculation enables us to convert from grams to moles. For 50 grams of CaCO\(_3\), we find: \[ \text{Number of moles} = \frac{50 \text{ g}}{100.09 \text{ g/mol}} = 0.5 \text{ mol} \]

Chemical Equation Balancing

Balancing chemical equations ensures the conservation of mass and atoms in a reaction. Each element must have the same number of atoms on both sides of the reaction. In our problem, the equation is: \[ \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \] Starting with reactants, we have 1 Ca, 1 C, and 3 O in CaCO\(_3\); and 2 H and 2 Cl in 2HCl. On the product side, we have 1 Ca in CaCl\(_2\), 1 C and 2 O in CO\(_2\), and 2 H and 1 O in H\(_2\)O. Notice all atoms balance perfectly across the equation. By balancing the equation first, we accurately determine the stoichiometric relationships used in later calculations.