The reaction between 27.2 g of HCI and 0.44 mol of manganese dioxide (MnO) yields manganese (I) chloride (MnCl), Cl₂, and H₂O as products. Answer the following questions regarding the reaction. a. Which reactant will be used up first? b. What is the theoretical yield of Cl₂ in grams? c. What is the percent yield if the actual yield is 11.67 g of Cl₂2
Answer (1)
a. HCl is the limiting reactant and will be used up first. b. The theoretical yield of Cl₂ is 13.23 g. c. The percent yield is 88.2%.[tex]________________________________[/tex]Let's go step by step to solve this stoichiometry problem.Given dataMass of HCl = 27.2 g Moles of MnO₂ = 0.44 molProducts: MnCl, Cl₂, and H₂OActual yield of Cl₂ = 11.67 g Step 1: Write the balanced chemical equation [tex]\[MnO_2 + 4HCl → MnCl_2 + Cl_2 + 2H_2O\][/tex]From the balanced equation: 1 mol of MnO₂ reacts with 4 mol of HCl Produces 1 mol of Cl₂ Step 2: Determine the limiting reactant Find the moles of HCl: Molar mass of HCl = 36.46 g/mol[tex]\[\frac{27.2 \text{ g HCl}}{36.46 \text{ g/mol}} = 0.746 \text{ mol HCl}\][/tex]Compare the mole ratio: The reaction needs 4 mol HCl per 1 mol MnO₂ The available ratio: [tex]\[\frac{0.746 \text{ mol HCl}}{4} = 0.186 \text{ mol MnO₂} \quad \text{(HCl can react with only 0.186 mol MnO₂)}\][/tex]Since we have 0.44 mol MnO₂ available but only enough HCl to react with 0.186 mol MnO₂, HCl is the limiting reactant.Step 3: Find the theoretical yield of Cl₂ From the balanced equation: 4 mol HCl produces 1 mol Cl₂ 0.746 mol HCl will produce: [tex]\[\frac{0.746}{4} = 0.1865 \text{ mol Cl₂}\][/tex]Molar mass of Cl₂ = 70.90 g/mol[tex]\[0.1865 \times 70.90 = 13.23 \text{ g Cl₂}\][/tex]Thus, the theoretical yield of Cl₂ is 13.23 g.Step 4: Calculate the percent yield [tex]\[\text{Percent yield} = \left(\frac{\text{actual yield}}{\text{theoretical yield}}\right) \times 100\][/tex][tex]\[= \left(\frac{11.67}{13.23}\right) \times 100\][/tex][tex]\[= 88.2\%\][/tex]Final Answersa. HCl is the limiting reactant and will be used up first. b. The theoretical yield of Cl₂ is 13.23 g. c. The percent yield is 88.2%.
