Consider the reaction shown in the following balanced equation:2 Cu2S (s) + 3 O2 (g) → 2 Cu2O (s) + 2 SO2 (g)An excess of Cu2S is added to 8.25 x 10−2 moles of O2.If all of the O2 reacts, the amount of SO2 that can theoretically be produced is Answer 1 Question 2 moles or Answer 2 Question 2 g. Molar mass of SO2 is 64.1 g mol−1.The reaction was performed in a lab and only 3.27 g of SO2 was produced. The percentage yield for the experiment was Answer 3 Question 2%.
Question
Consider the reaction shown in the following balanced equation:2 Cu2S (s) + 3 O2 (g) → 2 Cu2O (s) + 2 SO2 (g)An excess of Cu2S is added to 8.25 x 10−2 moles of O2.If all of the O2 reacts, the amount of SO2 that can theoretically be produced is Answer 1 Question 2 moles or Answer 2 Question 2 g. Molar mass of SO2 is 64.1 g mol−1.The reaction was performed in a lab and only 3.27 g of SO2 was produced. The percentage yield for the experiment was Answer 3 Question 2%.
Solution
To answer this question, we need to follow these steps:
Step 1: Identify the limiting reactant. In this case, we are told that there is an excess of Cu2S, so O2 is the limiting reactant.
Step 2: Use stoichiometry to determine the theoretical yield of SO2. From the balanced equation, we can see that 3 moles of O2 produce 2 moles of SO2. So, (8.25 x 10−2 moles of O2) x (2 moles of SO2 / 3 moles of O2) = 0.055 moles of SO2.
Step 3: Convert the moles of SO2 to grams using the molar mass. (0.055 moles of SO2) x (64.1 g of SO2 / 1 mole of SO2) = 3.53 g of SO2. This is the theoretical yield, so Answer 1 is 0.055 moles and Answer 2 is 3.53 g.
Step 4: Calculate the percentage yield. The percentage yield is the actual yield (what was actually produced in the lab) divided by the theoretical yield (what should have been produced according to the stoichiometry of the reaction), multiplied by 100%. So, (3.27 g / 3.53 g) x 100% = 92.64%. Therefore, Answer 3 is 92.64%.
Similar Questions
Consider the reaction shown in the following balanced equation:2 Cu2S (s) + 3 O2 (g) → 2 Cu2O (s) + 2 SO2 (g)You have 1.00 mole each of Cu2S and O2.How many moles of Cu2O can be produced if 1.00 mole of Cu2S reacts? Answer 1 Question 1How many moles of Cu2O can be produced if 1.00 mole of O2 reacts? Answer 2 Question 1The amount of Cu2O that can be produced is smaller from 1.00 mole of Answer 3 Question 1 than the other reagent. Therefore, Answer 4 Question 1 is the limiting reagent and Answer 5 Question 1 is the reagent in excess when 1.00 mole of each is available. Consider instead that you have 1.00 g each of Cu2S (molar mass of Cu2S is 159 g mol−1) and O2 (molar mass of O2 is 32.0 g mol−1).1.00 g of Cu2S is Answer 6 Question 1 moles of Cu2S and would produce Answer 7 Question 1 moles of Cu2O if all the Cu2S reacts. 1.00 g of O2 is Answer 8 Question 1 moles of O2 and would produce Answer 9 Question 1 moles of Cu2O if all the O2 reacts.The amount of Cu2O that can be produced is smaller from 1.00 g of Answer 10 Question 1 than the other reagent. Therefore, Answer 11 Question 1 is the limiting reagent and Answer 12 Question 1 is the reagent in excess when 1.00 g of each is available.
The law of conservation of mass states that:The mass of products is greater than the mass of reactantsThe mass of products is less than the mass of reactantsThe mass of products is equal to the mass of reactantsThe mass of products can be either greater than or less than the mass of reactants2How many grams of copper oxide are produced during the following reaction (assuming it is complete)?: 2Cu (6.35 g) + O2 (1.60 g) → 2CuO7.95 g4.75 g6.35 g15.9 g3In a chemical reaction, the mass of products is often lower than expected. There are several explanations for why this might be the case. Which of the following is not a possible explanation?Some of the reactants may have reacted in different unexpected waysThe mass of the reactants is greater than the mass of the productsThe reaction may not have finishedSome of the product may have been lost during removal4Which of the following is a balanced equation for the formation of water from hydrogen and oxygen?2H2 + O2 → H2O2H2 + 2O2 → 2H2O2H2 + O2 → 2H2OH2 + O2 → H2O25Which of the following is a balanced equation for the reaction of hydrogen and chlorine?H2 + Cl2 → HCl2H + 2Cl → 2HClH2 + Cl2 → 2HClH + Cl → HCl
Given the balanced chemical equation below:2 H2 + O2 --> 2 H2OHow many grams of water are produced when 64.0 g O2 gas reacts completely? You will need to calculate molar masses of O2 gas and H2O firstGroup of answer choices2.00 g4.00 g18.02 g72.1 g
2 is produced by the reaction of 0.4235 mol of CuCl2 according to the following equation: 2CuCl2+4KI→2CuI+4KCl+I22CuCl2+4KI→2CuI+4KCl+I2 .
2SO2+O2→2SO3 Rate of formation of SO3 according to the reaction is 1.6×10−3 kg min−1. Hence rate at which SO2 reacts is:
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.