\- Use the equation to calculate the following: $$ \mathrm{K}_{2} \mathrm{CO}_{3}(a q)+2 \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) \longrightarrow $$ \(2 \mathrm{KC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) (a) the moles of \(\mathrm{H}_{2} \mathrm{O}\) that can be obtained from \(25.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (b) the volume of \(0.210 \quad M \mathrm{~K}_{2} \mathrm{CO}_{3}\) needed to produce \(17.5 \mathrm{~mol} \mathrm{KC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (c) the volume of \(1.25 \mathrm{M} \mathrm{HC} \mathrm{H}_{2} \mathrm{H}_{2}\) needed to react with \(75.2 \mathrm{~mL} 0.750 \mathrm{M} \mathrm{K}_{2} \mathrm{CO}_{3}\) (d) the molarity \((M)\) of the \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) solution when \(10.15 \mathrm{~mL}\) react with \(18.50 \mathrm{~mL}\) of \(0.250 \mathrm{MK}_{2} \mathrm{CO}_{3}\) (e) the liters of \(\mathrm{CO}_{2}\) gas at STP produced by the reaction of \(105 \mathrm{~mL}\) of \(1.5 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (f) the liters of \(\mathrm{CO}_{2}\) gas at STP produced by the reaction of \(25.0 \mathrm{~mL}\) of \(0.350 M \mathrm{~K}_{2} \mathrm{CO}_{3}\) and \(25.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

Step by step solution

01

- Write Down Balanced Equation

Firstly, confirm the balanced chemical reaction given in the exercise: \[\mathrm{K}_{2}CO_{3}(aq) + 2\mathrm{HC}_{2}H_{3}O_{2}(aq) \rightarrow 2\mathrm{KC}_{2}H_{3}O_{2}(aq) + \mathrm{H}_{2}O(l) + \mathrm{CO}_{2}(g)\] This step is crucial as it will be used to determine the stoichiometric relationships between reactants and products.

02

- Calculate Moles of H2O (Part a)

To find the moles of \(\mathrm{H}_{2}\mathrm{O}\) that can be obtained from 25.0 mL of 0.150 M \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\), use the molarity-concentration equation: \[\text{Moles} = \text{Volume in L} \times \text{Molarity}\]Convert the volume from mL to L by dividing by 1000, then calculate the moles of \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\). Given the stoichiometry, each mole of \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) produces half a mole of \(\mathrm{H}_{2}\mathrm{O}\).

03

- Calculate Volume of K2CO3 (Part b)

To determine the volume of 0.210 M \(\mathrm{K}_{2}\mathrm{CO}_{3}\) needed to produce 17.5 moles of \(\mathrm{KC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\), use the stoichiometry from the balanced equation. Calculate the moles of \(\mathrm{K}_{2}\mathrm{CO}_{3}\) needed and then use the molarity equation to find the volume in liters.

04

- Calculate Volume of HC2H3O2 (Part c)

Find the volume of 1.25 M \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) required to react with 75.2 mL of 0.750 M \(\mathrm{K}_{2}\mathrm{CO}_{3}\). Start by determining the moles of \(\mathrm{K}_{2}\mathrm{CO}_{3}\) using the molarity-concentration equation, then use the stoichiometry to find the required moles and volume of \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\).

05

- Calculate Molarity of HC2H3O2 (Part d)

To compute the molarity of the \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) solution, use the volumes of the reacting solutions and their molarities. First calculate the moles of \(\mathrm{K}_{2}\mathrm{CO}_{3}\) based on the given volume and molarity. Then use the stoichiometry to find moles of \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) and ultimately its molarity by dividing by 10.15 mL converted to liters.

06

- Calculate Liters of CO2 at STP (Part e)

Determine the liters of \(\mathrm{CO}_{2}\) gas at STP produced by reaction of 105 mL of 1.5 M \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\). Calculate the moles of acid using its volume and molarity, then use stoichiometry to find the moles of \(\mathrm{CO}_{2}\). Apply the ideal gas law at STP to convert moles to volume in liters, using the fact that 1 mole of gas occupies 22.4 liters at STP.

07

- Calculate Liters of CO2 at STP (Part f)

To calculate the liters of \(\mathrm{CO}_{2}\) gas at STP produced from 25.0 mL of 0.350 M \(\mathrm{K}_{2}\mathrm{CO}_{3}\) and 25.0 mL of 0.250 M \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\), calculate the moles of each reactant. The limiting reactant will determine the maximum amount of \(\mathrm{CO}_{2}\) that can form. Use stoichiometry to find moles of \(\mathrm{CO}_{2}\) and then convert to liters at STP using the ideal gas law.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reaction Equations

The accurate representation of chemical reactions is fundamental in chemistry, and it's achieved through balanced chemical reaction equations. These equations succinctly convey the proportional relationship between reactants and products. The law of conservation of mass dictates that for an equation to be balanced, the number of atoms of each element must be the same on both sides of the reaction.

For example, in the mentioned exercise, \(\mathrm{K}_{2}CO_{3}(aq) + 2\mathrm{HC}_{2}H_{3}O_{2}(aq) \rightarrow 2\mathrm{KC}_{2}H_{3}O_{2}(aq) + \mathrm{H}_{2}O(l) + \mathrm{CO}_{2}(g)\) demonstrates the reaction between potassium carbonate and acetic acid to produce potassium acetate, water, and carbon dioxide. To ensure that calculations based on this equation are accurate, one must confirm that the equation is balanced, meaning each type of atom has the same count on both sides of the equation.

Molarity Calculations

Molarity, or molar concentration, is a measure of the concentration of a solute in a solution. It is defined as the moles of solute divided by the liters of solution. The molarity formula is \( M = \text{moles solute} / \text{liters solution}\).

This concept is crucial when mixing chemicals in a lab or interpreting the results of a chemical reaction. For instance, to find out how much product can be made from a certain volume of reactant, we must convert volumes to moles using molarity. In practice, volume must be in liters, so a common step is to convert milliliters to liters by dividing by 1000, as seen in the exercise.

Limiting Reactant Analysis

When performing chemical reactions, there's often a reactant that runs out first, called the limiting reactant, which in turn stops the reaction and determines the maximum amount of product that can be made. Identifying the limiting reactant is essential for calculating theoretical yields and can be done through stoichiometry.

To find the limiting reactant, compute the moles of each reactant and compare them based on the reaction's stoichiometry. The reactant that makes less product (or runs out first) is the limiting reactant. This ties directly to the exercise where two reactants are combined, and knowing their molarities and volumes allows one to identify the limiting reactant and then calculate the amount of product produced at the end.

Ideal Gas Law Application

The ideal gas law, given by \(PV = nRT\), where P is pressure, V is volume, n is moles of gas, R is the gas constant, and T is temperature, describes the behavior of an ideal gas. At standard temperature and pressure (STP: 0°C and 1 atm), 1 mole of any ideal gas occupies 22.4 liters.

This law allows us to calculate one property of a gas if we know the others. In the exercise, the ideal gas law is applied to convert moles of \(\mathrm{CO}_{2}\) gas into liters at STP. Knowing that conditions at STP simplifies calculations, as one can use the 22.4 liters per mole relationship without adjusting R or T to find the volume the gas will occupy at these standard conditions.

Mole-to-Mole Ratio

In chemistry, reactions proceed according to specific mole-to-mole ratios, which are derived from the balanced chemical equation. The coefficients in a balanced equation provide the ratio of reactants to products, which is essential for stoichiometric calculations.

For example, the balanced equation \(2\mathrm{H}_{2} + \mathrm{O}_{2} \rightarrow 2\mathrm{H}_{2}\mathrm{O}\) indicates that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water. This 2:1:2 ratio is key when calculating quantities of reactants or products, just like it was necessary to apply this principle to solve the parts of our original exercise that involve translating the amount of one substance into the equivalent amount of another via the reaction’s inherent ratios.