Question

A quantity of \(0.560 \mathrm{~g}\) of \(\mathrm{KOH}\) is added to \(25.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). Excess \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is then added to the solution. What mass (in grams) of \(\mathrm{CO}_{2}\) is formed?

Short answer

The mass of \(\mathrm{CO}_2\) that is formed is approximately \(0.660\) g.

Step-by-step solution

Calculate moles of KOH

First, calculate the moles of \(\mathrm{KOH}\). The molar mass of \(\mathrm{KOH}\) is approximately \(56.11 \mathrm{~g/mol}\). So the moles of \(\mathrm{KOH}\) are \(\frac{0.560 ~g}{56.11 ~g/mol} \approx 0.01 \mathrm{~mol}\) .

Calculate moles of HCl initially and after reaction with KOH

Initially, \(25.0\) \(\mathrm{mL}\) of \(1.00 \) \(\mathrm{M} \mathrm{HCl}\) corresponds to \(1.00 \mathrm{M} * 0.025 ~L = 0.025 \mathrm{~mol}\). All the \(\mathrm{KOH}\) reacts with the \(\mathrm{HCl}\), using up \(0.01 \mathrm{~mol}\) of \(\mathrm{HCl}\). So the moles of \(\mathrm{HCl}\) remaining are \(0.025 \mathrm{~mol} - 0.01 \mathrm{~mol} = 0.015 \mathrm{~mol}\).

Calculate moles of CO2 formed

Each mole of \( \mathrm{HCl} \) reacts with \( \mathrm{Na}_2 \mathrm{CO}_3 \) to produce one mole of \( \mathrm{CO}_2 \). So, \( 0.015 \)moles of \( \mathrm{HCl} \) will produce \( 0.015 \) moles of \( \mathrm{CO}_2 \) .

Calculate mass of CO2 formed

Finally, convert moles of \( \mathrm{CO}_2 \) to grams using its molar mass (around \(44.01 \)g/mol): \(0.015 \)mol * \(44.01 \)g/mol = 0.660 g.

Key concepts

Mole Calculations

Moles are an essential unit in chemistry that help us transition between the mass of a substance and the number of atoms or molecules. In this exercise, mole calculations allow us to understand the quantities of reactants and products involved.
To find moles, we divide the mass of a substance by its molar mass (mass of one mole of the substance in grams).

• For \(\mathrm{KOH}\), with a given mass of \(0.560 \, \mathrm{g}\) and a molar mass of approximately \(56.11 \, \mathrm{g/mol}\), the moles are calculated as \(\frac{0.560 \, \mathrm{g}}{56.11 \, \mathrm{g/mol}} \approx 0.01 \, \mathrm{mol}\).
• Similarly, to find moles of an initial solution, multiply its concentration by volume. For \(\mathrm{HCl}\), \(1.00 \, \mathrm{M}\) concentration and \(0.025 \, \mathrm{L}\) volume yields \(0.025 \, \mathrm{mol}\).

With moles, you can track how much of a substance is present and how far a reaction has progressed in terms of amount, making it invaluable for chemical problem-solving.

Acid-Base Reaction

Acid-base reactions involve acids and bases neutralizing each other, producing water and a salt. In our given problem, \(\mathrm{KOH}\), a base, reacts with \(\mathrm{HCl}\), an acid. This type of reaction is fundamental in chemistry and showcases a simple neutralization case.

• Reaction: The reaction between \(\mathrm{KOH}\) and \(\mathrm{HCl}\) forms \(\mathrm{KCl}\) and water: \(\mathrm{KOH + HCl} \rightarrow \mathrm{KCl + H_2O}\).
• Neutralization: In neutralization, the number of moles of \(\mathrm{H^+}\) ions from the acid equals moles of \(\mathrm{OH^-}\) ions from the base. Thus, \(0.01 \, \mathrm{mol}\) of \(\mathrm{KOH}\) neutralizes \(0.01 \, \mathrm{mol}\) of \(\mathrm{HCl}\).

Only excess \(\mathrm{HCl}\) remains, which is important for subsequent reactions. So, after neutralizing with \(\mathrm{KOH}\), \(0.015 \, \mathrm{mol}\) of \(\mathrm{HCl}\) is left to react further.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It is crucial for understanding how much of each substance is involved and formed in a given reaction. Here, stoichiometry guides us through the step by step solving of how much \(\mathrm{CO}_2\) forms.

• Guiding Principles: Use balanced chemical equations to understand ratios. In this problem: \(\mathrm{HCl}\) reacts with \(\mathrm{Na_2CO_3}\) to form \(\mathrm{CO_2}\). Each mole of \(\mathrm{HCl}\) corresponds to one mole of \(\mathrm{CO_2}\).
• Application: Since \(0.015 \, \mathrm{mol}\) of \(\mathrm{HCl}\) remains, it produces \(0.015 \, \mathrm{mol}\) of \(\mathrm{CO_2}\).

To determine \(\mathrm{CO_2}\) mass: multiply moles by molar mass of \(\mathrm{CO_2}\) (around \(44.01 \, \mathrm{g/mol}\)), calculating \(0.015 \, \text{mol} \times 44.01 \, \mathrm{g/mol} = 0.660 \, \mathrm{g}\). This final step, rooted in stoichiometry, reveals the mass of carbon dioxide formed, bringing the solution full circle.