3) Medicinal "ether" is produced when ethyl alcohol is treated with an acid. How many grams of medicinal "ether" (C، $\left.\mathrm{C}_{12} \mathrm{O}\right)$ would you produce with $50.0 \mathrm{~g}$ of ethyl alcohol $\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{O}\right)$ if the percent yield is $85 \%$ ?
$2 \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} \rightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}$
4) $\mathrm{A} 8.15 \mathrm{~g}$ sample of hydrogen peroxide $\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)$ decomposes to form water and oxygen. The temperature and pressures conditions in the lab were 21.2 eC and 761.4 torr, respectively. The oxygen gas is collected over a sample of water at $21.20 \mathrm{C}$; the vapor pressure of water at that temperature is $18.9 \mathrm{torm}$. When the water level inside and outside of the tube is equal the volume of of gas is recurded as 176.23 $\mathrm{mL}$.
a) Write and balance the equation for the decomposition reaction
b) What is the partial pressure of the oxygen gas?
c) How many moles of oxygen gas were produced?
d) What was the mass of hydrogen peroxide that reacted?
c) What is the percent by mass of the hydrogen peroxide from the original sample?
5) A $25.0 \mathrm{~mL}$ sample of $0.25 \mathrm{M}$ potassium carbonate $\left(\mathrm{K}_{2} \mathrm{CO}_{2}\right)$ solution is added to $30.0 \mathrm{~mL}$ of a $0.40 \mathrm{M}$ barium nitrate $\left(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\right)$ solution. What is the concentration of the excess metal ion after the precipitation reaction is complete?

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Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Calculate the molar mass of ethyl alcohol (C2H5OH)**: To find the mass of medicinal ether produced, we first need to calculate the molar mass of ethyl alcohol. The molar mass of C2H5OH is calculated as follows: $2 imes 12.01 ext{ g/mol} + 6 imes 1.008 ext{ g/mol} + 16.00 ext{ g/mol} = 46.08 ext{ g/mol}$. ‖ ‖ 2 ⋮ **Determine moles of ethyl alcohol**: Using the molar mass, we can find the number of moles of ethyl alcohol in 50.0 g. $\frac{50.0 ext{ g}}{46.08 ext{ g/mol}} = 1.085 ext{ moles of C2H5OH}$. ‖ ‖ 3 ⋮ **Use stoichiometry to find moles of ether produced**: According to the balanced equation, 2 moles of ethyl alcohol produce 1 mole of ether. Therefore, $1.085 ext{ moles of C2H5OH} imes \frac{1 ext{ mole of ether}}{2 ext{ moles of C2H5OH}} = 0.5425 ext{ moles of ether}$. ‖ ‖ 4 ⋮ **Calculate the theoretical yield of ether**: The molar mass of medicinal ether (C4H10O) is $4 imes 12.01 ext{ g/mol} + 10 imes 1.008 ext{ g/mol} + 16.00 ext{ g/mol} = 74.12 ext{ g/mol}$. The theoretical yield is $0.5425 ext{ moles} imes 74.12 ext{ g/mol} = 40.21 ext{ g}$. ‖ ‖ 5 ⋮ **Calculate the actual yield based on percent yield**: The actual yield is found by multiplying the theoretical yield by the percent yield. $40.21 ext{ g} imes \frac{85}{100} = 34.18 ext{ g}$. ‖ ∻3 Answer∻ ⚹ 34.18 g ⚹ ∻Key Concept∻ ⚹ Percent yield is used to determine the actual yield of a product in a chemical reaction based on the theoretical yield and the efficiency of the reaction. ⚹ ∻Explanation∻ ⚹ The actual yield of medicinal ether is calculated by first determining the theoretical yield from the stoichiometry of the balanced equation and then applying the percent yield to find the mass of ether that would be produced in practice. ⚹ ∻Solution∻ ‖ 1 ⋮ **Write and balance the decomposition reaction**: The balanced chemical equation for the decomposition of hydrogen peroxide is $2 ext{H}_2 ext{O}_2 ightarrow 2 ext{H}_2 ext{O} + ext{O}_2$. ‖ ‖ 2 ⋮ **Calculate the partial pressure of oxygen gas**: The total pressure is the sum of the partial pressures of oxygen and water vapor. The partial pressure of oxygen is the total pressure minus the vapor pressure of water. $761.4 ext{ torr} - 18.9 ext{ torr} = 742.5 ext{ torr}$. ‖ ‖ 3 ⋮ **Convert the partial pressure to atm and calculate moles of oxygen gas**: Using the ideal gas law, $PV = nRT$, where $P$ is the pressure in atm, $V$ is the volume in liters, $R$ is the gas constant (0.0821 L·atm/mol·K), and $T$ is the temperature in Kelvin. First, convert the pressure to atm ($742.5 ext{ torr} imes \frac{1 ext{ atm}}{760 ext{ torr}} = 0.977 ext{ atm}$) and the volume to liters ($176.23 ext{ mL} imes \frac{1 ext{ L}}{1000 ext{ mL}} = 0.17623 ext{ L}$). The temperature in Kelvin is $21.2^\circ ext{C} + 273.15 = 294.35 ext{ K}$. Now, solve for $n$: $n = \frac{PV}{RT} = \frac{0.977 ext{ atm} imes 0.17623 ext{ L}}{0.0821 ext{ L·atm/mol·K} imes 294.35 ext{ K}} = 0.0073 ext{ moles of O}_2$. ‖ ‖ 4 ⋮ **Calculate the mass of hydrogen peroxide that reacted**: Using the stoichiometry from the balanced equation, 2 moles of H2O2 produce 1 mole of O2. Therefore, the moles of H2O2 that reacted are $0.0073 ext{ moles of O}_2 imes \frac{2 ext{ moles of H}_2 ext{O}_2}{1 ext{ mole of O}_2} = 0.0146 ext{ moles of H}_2 ext{O}_2$. The molar mass of H2O2 is $2 imes 1.008 ext{ g/mol} + 2 imes 16.00 ext{ g/mol} = 34.016 ext{ g/mol}$. The mass of H2O2 that reacted is $0.0146 ext{ moles} imes 34.016 ext{ g/mol} = 0.4966 ext{ g}$. ‖ ‖ 5 ⋮ **Calculate the percent by mass of hydrogen peroxide in the original sample**: The percent by mass is the mass of H2O2 that reacted divided by the original mass of the sample, multiplied by 100. $\frac{0.4966 ext{ g}}{8.15 ext{ g}} imes 100 = 6.09\%$. ‖ ∻4 Answer∻ ⚹ a) 2 H2O2 → 2 H2O + O2 b) 742.5 torr c) 0.0073 moles d) 0.4966 g e) 6.09% ⚹ ∻Key Concept∻ ⚹ The ideal gas law relates the pressure, volume, temperature, and number of moles of a gas, and can be used to calculate the amount of reactants or products in a chemical reaction. ⚹ ∻Explanation∻ ⚹ The partial pressure of oxygen is found by subtracting the vapor pressure of water from the total pressure. The moles of oxygen are calculated using the ideal gas law, and this information is used to determine the mass of hydrogen peroxide that reacted and its percent by mass in the original sample. ⚹ ∻Solution∻ ‖ 1 ⋮ **Calculate moles of potassium carbonate and barium nitrate**: For potassium carbonate, $0.025 ext{ L} imes 0.25 ext{ M} = 0.00625 ext{ moles}$. For barium nitrate, $0.030 ext{ L} imes 0.40 ext{ M} = 0.012 ext{ moles}$. ‖ ‖ 2 ⋮ **Write the balanced equation for the precipitation reaction**: The balanced chemical equation is $K_2CO_3 + Ba(NO_3)_2 ightarrow BaCO_3 \downarrow + 2 KNO_3$. ‖ ‖ 3 ⋮ **Determine the limiting reactant**: From the stoichiometry of the balanced equation, 1 mole of K2CO3 reacts with 1 mole of Ba(NO3)2. Since there are fewer moles of K2CO3, it is the limiting reactant. ‖ ‖ 4 ⋮ **Calculate the concentration of the excess metal ion**: All of the K2CO3 will react, leaving some Ba(NO3)2 unreacted. The remaining moles of Ba(NO3)2 are $0.012 ext{ moles} - 0.00625 ext{ moles} = 0.00575 ext{ moles}$. The total volume of the solution after mixing is $25.0 ext{ mL} + 30.0 ext{ mL} = 55.0 ext{ mL}$. The concentration of the excess Ba^2+ ions is $\frac{0.00575 ext{ moles}}{0.055 ext{ L}} = 0.1045 ext{ M}$. ‖ ∻5 Answer∻ ⚹ 0.1045 M ⚹ ∻Key Concept∻ ⚹ In a precipitation reaction, the limiting reactant determines the amount of product formed, and the excess reactant remains in solution after the reaction is complete. ⚹ ∻Explanation∻ ⚹ The concentration of the excess metal ion after the precipitation reaction is calculated by determining the limiting reactant, finding the remaining moles of the excess reactant, and dividing by the total volume of the solution. ⚹◊What is the chemical equation for the decomposition of hydrogen peroxide into water and oxygen?⍭ Generate me a similar question◊

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