Question

The oxides of Group 2 A metals (symbolized by M here) react with carbon dioxide according to the following reaction: $$ \mathrm{MO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{MCO}_{3}(s) $$ A 2.85-g sample containing only \(\mathrm{MgO}\) and \(\mathrm{CuO}\) is placed in a 3.00-L container. The container is filled with \(\mathrm{CO}_{2}\) to a pressure of 740 . torr at \(20 .{ }^{\circ} \mathrm{C}\). After the reaction has gone to completion, the pressure inside the flask is 390 . torr at \(20 .{ }^{\circ} \mathrm{C}\). What is the mass percent of \(\mathrm{MgO}\) in the mixture? Assume that only the \(\mathrm{MgO}\) reacts with \(\mathrm{CO}_{2}\).

Short answer

The mass percent of \(\mathrm{MgO}\) in the mixture is 59.6%.

Step-by-step solution

Calculate moles of \(\mathrm{CO}_{2}\) consumed during the reaction

Using the Ideal Gas Law formula, \(PV = nRT\), we will find the initial moles of \(\mathrm{CO}_{2}\) and then the final moles after the reaction. From there, we can find the amount consumed during the reaction. Initial moles of \(\mathrm{CO}_{2}\), \(n_{i} = \frac{P_{i}V}{RT}\) Final moles of \(\mathrm{CO}_{2}\), \(n_{f} = \frac{P_{f}V}{RT}\) Change in moles of \(\mathrm{CO}_{2}\), \(\Delta n = n_{i} - n_{f}\) Note: \(R = 62.364 \frac{L * torr}{K * mol}\).

Calculate the mass of \(\mathrm{MgO}\) reacted

Now, we will use stoichiometry to calculate the mass of \(\mathrm{MgO}\) reacted. From the balanced equation, we can see that 1 mole of \(\mathrm{MgO}\) reacts with 1 mole of \(\mathrm{CO}_{2}\), so: Mass of \(\mathrm{MgO}\) reacted = (moles of \(\mathrm{CO}_{2}\) consumed) * (molar mass of \(\mathrm{MgO}\))

Calculate the mass percent of \(\mathrm{MgO}\) in the mixture

Now that we have the mass of \(\mathrm{MgO}\) reacted, we can find its mass percentage in the mixture. Mass percent of \(\mathrm{MgO}\) = \(\frac{\text{Mass of MgO reacted}}{\text{Total mass of mixture}} \times 100\) Let's perform the calculations using the given information: 1. Calculate initial moles of \(\mathrm{CO}_{2}\): \(n_{i} = \frac{P_{i}V}{RT} = \frac{740\,\text{torr} \cdot 3.00\,\text{L}}{(20 + 273.15)\,\text{K} \cdot 62.364\,\frac{\text{L}\cdot\text{torr}}{\text{K}\cdot\text{mol}}} = 0.0896\,\text{mol}\) 2. Calculate final moles of \(\mathrm{CO}_{2}\): \(n_{f} = \frac{P_{f}V}{RT} = \frac{390\,\text{torr} \cdot 3.00\,\text{L}}{(20 + 273.15)\,\text{K} \cdot 62.364\,\frac{\text{L}\cdot\text{torr}}{\text{K}\cdot\text{mol}}} = 0.04740\,\text{mol}\) 3. Calculate the change in moles of \(\mathrm{CO}_{2}\): \(\Delta n = n_{i} - n_{f} = 0.0896\,\text{mol} - 0.04740\,\text{mol} = 0.0422\,\text{mol}\) 4. Calculate the mass of \(\mathrm{MgO}\) reacted: Mass of \(\mathrm{MgO}\) reacted = (0.0422 mol) * (40.305 g/mol \(\mathrm{MgO}\)) = 1.70 g 5. Calculate the mass percent of \(\mathrm{MgO}\) in the mixture: Mass percent of \(\mathrm{MgO}\) = \(\frac{1.70\,\text{g}}{2.85\,\text{g}} \times 100 = 59.6\%\) So, the mass percent of \(\mathrm{MgO}\) in the mixture is 59.6%.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that relates the pressure, volume, temperature, and number of moles of a gas through the equation \(PV = nRT\). In this equation, \(P\) represents the pressure of the gas, \(V\) stands for volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.
This equation is an essential tool for solving problems involving gases, especially in closed systems where these variables interact.

• Temperature must be converted to Kelvin by adding 273.15 to the Celsius value.
• The gas constant \(R\) has different values based on the units of pressure; here, it is used as 62.364 \( L \cdot torr / K \cdot mol \).
• This law assumes the gas behaves ideally, where interactions between molecules are negligible.

In the provided exercise, this law helps calculate the initial and final moles of \(\mathrm{CO}_2\) by rearranging the formula to \(n = \frac{PV}{RT}\).
Understanding this application of the Ideal Gas Law is crucial for determining changes in moles due to gas reactions.

Gas Reactions

Gas reactions often involve a transformation where reactant gases are converted into other substances, sometimes resulting in a change in pressure or volume.
In this exercise, the focus is on a reaction between solid \(\mathrm{MO}\) compounds and \(\mathrm{CO}_2\) gas. The reaction proceeds according to the equation:

• \(\mathrm{MO}(s) + \mathrm{CO}_2(g) \longrightarrow \mathrm{MCO}_3(s)\)

This shows that a gaseous reactant, \(\mathrm{CO}_2\), is reacting with a solid \(\mathrm{MO}\) to form another solid, \(\mathrm{MCO}_3\).

• Reactions where there is a decrease in the number of gas molecules typically result in a reduction in gas pressure, as seen in the exercise.
• Reactions involving gases can be monitored by measuring changes in pressure, using the Ideal Gas Law to relate these changes back to the number of moles reacting.

It's essential to understand that, in this specific reaction, only one component of the solid mixture is reactive. This behavior allows for determining the quantity of \(\mathrm{MgO}\) using pressure change measurements.

Mass Percent Calculation

Mass percent calculation is a method to determine the proportion of a particular compound in a mixture, expressed as a percentage by mass.
It helps in understanding the composition of mixtures and is a common calculation in chemistry.

• It is done using the formula \(\frac{\text{Mass of Component}}{\text{Total Mass of Mixture}} \times 100\). This gives the percentage of the component in the mixture.

In our problem, the mass percent of \(\mathrm{MgO}\) in a mixture that also contains \(\mathrm{CuO}\) is calculated after determining the mass of \(\mathrm{MgO}\) that reacted.

• It involves first calculating the mass of the \(\mathrm{MgO}\) that underwent reaction by using the stoichiometry of the reaction and moles of \(\mathrm{CO}_2\) consumed.
• Then applying the mass percent formula to find that \(59.6\%\) of the initial mixture's mass is due to \(\mathrm{MgO}\).

Understanding mass percent is crucial when analyzing mixtures to identify the amount of each component.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds, leading to the formation of new substances.
They are usually represented by chemical equations, which show the reactants transforming into products, as in the given equation:

• \(\mathrm{MO}(s) + \mathrm{CO}_2(g) \rightarrow \mathrm{MCO}_3(s)\)

In the context of this problem, note the following key points:

• The stoichiometry of a balanced chemical equation reveals the ratio in which reactants combine and products are formed.
• This reaction is particularly important for showing how changes in one reactant can predictably alter the product yield.

In the specific scenario given, only \(\mathrm{MgO}\) reacts with \(\mathrm{CO}_2\) to form \(\mathrm{MgCO}_3\), highlighting the selective nature of chemical reactivity.
Understanding such reactions is fundamental for predicting the outcomes of chemical processes and analyzing how different conditions affect reactions.