Question

What volume of \(3.44 \mathrm{M} \mathrm{HCl}\) will react with \(5.33 \mathrm{~mol}\) of \(\mathrm{CaCO}_{3}\) ? \(2 \mathrm{HCl}+\mathrm{CaCO}_{3} \rightarrow \mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\).

Short answer

3.10 liters of 3.44 M HCl are needed to react with 5.33 moles of CaCO₃.

Step-by-step solution

Write the Balanced Chemical Equation

The balanced chemical equation for the reaction is given as follows: \[ 2 \mathrm{HCl} + \mathrm{CaCO}_{3} \rightarrow \mathrm{CaCl}_{2} + \mathrm{H}_{2} \mathrm{O} + \mathrm{CO}_{2} \]This equation shows that 2 moles of HCl are required for 1 mole of CaCO₃.

Determine Mole Ratio

From the balanced equation, we can see that the mole ratio of \( \mathrm{HCl} \) to \( \mathrm{CaCO}_{3} \) is 2:1. This means 2 moles of HCl are needed to react with 1 mole of CaCO₃.

Calculate Moles of HCl Needed

Since the mole ratio is 2:1 and you have 5.33 moles of \( \mathrm{CaCO}_{3} \), you need:\[ 2 \times 5.33 = 10.66 \text{ moles of } \mathrm{HCl} \]So, 10.66 moles of HCl are required.

Calculate Volume of HCl Solution

To find the volume of the \(3.44 \text{ M HCl }\) solution needed, use the formula:\[ \text{Molarity (} M\text{)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]Re-arrange the formula to solve for volume:\[ \text{Volume } = \frac{10.66 \text{ moles}}{3.44 \text{ M}} \]\[ \text{Volume } = 3.10 \text{ liters} \]Thus, 3.10 liters of the HCl solution are required.

Key concepts

Mole Ratio

The concept of mole ratio is pivotal in understanding chemical reactions. A mole ratio is defined as the ratio of the amounts in moles of any two compounds involved in a chemical reaction. This ratio comes from the coefficients of the substances in the balanced chemical equation.

• Consider the reaction of hydrochloric acid (HCl) with calcium carbonate (CaCO₃).
• The balanced equation is: \( 2 \text{ HCl} + \text{ CaCO}_3 \rightarrow \text{ CaCl}_2 + \text{ H}_2\text{O} + \text{ CO}_2 \)
• The coefficients indicate that 2 moles of HCl react with 1 mole of CaCO₃.

Understanding this mole ratio is essential for calculating how much of a reactant is needed or how much product will form in a reaction. If you start with 5.33 moles of CaCO₃, you will need 2 times this amount of HCl, which is 10.66 moles, because of the 2:1 mole ratio.

Balanced Chemical Equation

A balanced chemical equation is crucial because it reflects the conservation of mass, where the number of atoms of each element is the same on both sides of the equation. This ensures that the law of conservation of matter is followed.

• In the equation for the reaction between HCl and CaCO₃: \(2 \text{ HCl} + \text{ CaCO}_3 \rightarrow \text{ CaCl}_2 + \text{ H}_2\text{O} + \text{ CO}_2\), each type of atom in the reactants equals the number of atoms in the products.
• This balance shows that no atoms are lost or gained, only rearranged.
• Properly balancing an equation is necessary for accurately predicting the amounts of substances consumed and produced in a reaction.

Before any stoichiometric calculations, make sure the chemical equation is balanced. This step sets the stage for determining the mole ratios used in further calculations.

Molarity

Molarity is a way to express the concentration of a solution, specifically indicating how many moles of solute are present in one liter of solution. This concept is vital for understanding how much of a solution is needed to react completely with a given amount of reactant.

• The molarity (M) of HCl solution used is given as 3.44 M.
• This means there are 3.44 moles of HCl in every liter of solution.
• To find out the volume of the solution needed, use the formula: \( M = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \).

If you need 10.66 moles of HCl, set up the equation to find the volume: \( \text{Volume} = \frac{10.66 \text{ moles}}{3.44 \text{ M}} = 3.10 \text{ liters} \). This calculation provides the precise volume of the solution necessary for the reaction, ensuring complete reaction with the given moles of CaCO₃.