Question

A seashell that is composed largely of calcium carbonate reacts with a solution of HCl. As a result, 1500 \(\mathrm{mL}\) of dry \(\mathrm{CO}_{2}\) gas at STP is produced. The other products are CaCl, and \(\mathrm{H}_{2} \mathrm{O}\) . a. How many grams of \(\mathrm{CaCO}_{3}\) are consumed in the reaction? b. What volume of 2.00 \(\mathrm{M}\) HCl solution is used in this reaction?

Short answer

a) 6.71 g \text{CaCO}_{3}. b) 67 \text{mL} \text HCl.

Step-by-step solution

- Write the Balanced Chemical Equation

First, write the balanced chemical equation for the reaction of calcium carbonate (\text{CaCO}_{3}) with hydrochloric acid (\text{HCl}). $$\text{CaCO}_{3} + 2\text{HCl} \rightarrow \text{CaCl}_{2} + \text{CO}_{2} + \text{H}_{2}O$$

- Convert Volume of \text{CO}_{2} Gas to Moles

Use the molar volume of a gas at STP (22.4 \text{L/mol}) to convert the given volume of \text{CO}_{2} gas (1500 \text{mL} or 1.5 \text{L}) to moles. $$\text{Moles of CO}_{2} = \frac{1.5 \text{L}}{22.4 \text{L/mol}} = 0.067 \text{ mol}$$

- Determine Moles of \text{CaCO}_{3}

Look at the stoichiometry of the reaction. 1 mole of \text{CaCO}_{3} produces 1 mole of \text{CO}_{2}. Therefore, the moles of \text{CaCO}_{3} consumed is the same as the moles of \text{CO}_{2} produced: $$\text{Moles of CaCO}_{3} = 0.067 \text{ mol}$$

- Calculate Mass of \text{CaCO}_{3}

Convert moles of \text{CaCO}_{3} to grams using its molar mass (100.09 \text{ g/mol}). $$\text{Mass of CaCO}_{3} = 0.067 \text{ mol} \times 100.09 \text{ g/mol} = 6.71 \text{ g}$$

- Find Volume of HCl Solution

Use the stoichiometry to determine moles of HCl required. From the equation, 2 moles of HCl react with 1 mole of \text{CaCO}_{3}. Thus, $$\text{Moles of HCl} = 2 \times 0.067 \text{ mol} = 0.134 \text{ mol}$$Then, convert moles of HCl to volume using the concentration of HCl (2.00 \text{M}):$$\text{Volume of HCl} = \frac{0.134 \text{ mol}}{2.00 \text{ mol/L}} = 0.067 \text{ L} = 67 \text{ mL}$$

Key concepts

Balanced Chemical Equations

Balanced chemical equations are crucial for understanding stoichiometry problems. In a balanced equation, the number of atoms for each element is the same on both the reactant and product sides. This balance maintains the law of conservation of mass. For example, consider the reaction of calcium carbonate (\(CaCO_3\)) with hydrochloric acid (\(HCl\)). The balanced equation is:\[CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O\]Here, each side has one calcium atom, three oxide atoms, one carbonate atom, and two hydrogen atoms. Balancing chemical equations ensures the correct proportions of reactants and products.

Molar Volume Calculation

Molar volume is the volume one mole of gas occupies, typically measured at standard temperature and pressure (STP). At STP, one mole of any ideal gas occupies 22.4 liters. This concept is key when converting gas volumes to moles. For instance, in the problem, 1500 mL of \(CO_2\) gas is produced. Converting to liters gives 1.5 L. Using the molar volume of 22.4 L/mol, the calculation becomes:\[\text{Moles of } CO_2 = \frac{1.5 \text{ L}}{22.4 \text{ L/mol}} = 0.067 \text{ mol}\]This process allows us to use the measured gas volume to determine the number of moles.

Mole-to-Mole Conversion

Mole-to-mole conversion uses the coefficients from a balanced chemical equation to relate the amounts of reactants and products. These coefficients tell us the ratio in which substances react or are produced. In the given problem, the balanced equation indicates that 1 mole of \(CaCO_3\) generates 1 mole of \(CO_2\). So, if we produce 0.067 moles of \(CO_2\), then the same amount of \(CaCO_3\) was consumed:\[\text{Moles of } CaCO_3 = 0.067 \text{ mol}\]Understanding these ratios simplifies how we translate between different components of the reaction.

Molar Mass Calculation

Molar mass helps convert between grams and moles for a given substance. The molar mass of a compound is the sum of the masses of its constituent elements. For \(CaCO_3\), this is:\[\text{Ca} = 40.08 \text{ g/mol}, \text{C} = 12.01 \text{ g/mol}, \text{O} = 3 \times 16.00 \text{ g/mol} = 48.00 \text{ g/mol}\]Adding these values gives the molar mass:\[100.09 \text{ g/mol}\]Using this, we calculated the mass of \(CaCO_3\) consumed:\[0.067 \text{ mol} \times 100.09 \text{ g/mol} = 6.71 \text{ g}\]Accurate molar mass calculation is essential for translating between different quantity units in chemistry.

Solution Concentration

Solution concentration, typically given in molarity (M), represents the number of moles of solute per liter of solution. It is vital for calculating how much reactant is needed. In our problem, the concentration of \(HCl\) is 2.00 M. First, we determined the moles of \(HCl\) required using the balanced equation: 2 moles of \(HCl\) for every 1 mole of \(CaCO_3\):\[2 \times 0.067 \text{ mol} = 0.134 \text{ mol}\]Then, we used the concentration to find the volume:\[\text{Volume of } HCl = \frac{0.134 \text{ mol}}{2.00 \text{ mol/L}} = 0.067 \text{ L} = 67 \text{ mL}\]Accurate knowledge of solution concentration ensures we use the correct amounts in reactions.