**Reaction of Benzene with Bromine to Form Bromobenzene**A student reacts benzene (C₆H₆) with bromine (Br₂) to prepare bromobenzene (C₆H₅Br) and hydrobromic acid (HBr).**Chemical Equation:**C₆H₆ + Br₂ → C₆H₅Br + HBr**Problem Statement:**1. What is the theoretical yield of bromobenzene in this reaction when 34.8 g of benzene reacts with 57.8 g of bromine? - [ ] grams2. Which chemical is the limiting reactant? - [Select an answer from the dropdown]3. If the actual yield of bromobenzene was 38.5 g, what was the percent yield? - [ ] %**Interactive Elements:**- Dropdown menu to select the limiting reactant.- Input box to enter the theoretical yield in grams.- Input box to enter the percent yield. - **Action Button:** - Submit Question

Given: Mass of benzene= 34.8 g Mass of bromine = 57.8 g Reaction is: C6H6 +Br2 -------> C6H5Br +HBrMoles of benzene=mass of benzeneMolar mass of benzene=34.8 g78.11 g/mol=0.4455 molMoles of Bromine=mass of BromineMolar mass of Bromine=57.8 g159.808 g/mol=0.3617 molMoles of Benzene=0.4455 molMoles of Bromine=0.3617 molNow according to the balanced equation, Moles of benzene is equal to the moles of bromine.But here, moles of Bromine are less than that of moles of Benzene.So, Bromine act as a LIMITING REACTANT.Theoretical yield of bromobenzene can be calculated as: Since Bromine is the limiting reagent, so the amount of bromobenzene formed will depend on the initial amount of the benzene. According to the balanced equation: Moles of Bromine= Moles of Bromobenzene This implies, Moles of bromobenzene = 0.3617 mol Theoretical yield of bromobenzene = Moles x Molar mass of bromobenzene = 0.3617 mol x 157.01 g/mol =56.79 g Theoretical yield of bromobenzene = 56.79 g To Calculate the percent yield Percent yield =Theoretical yield- Actual yieldTheoretical yield× 100%Percent yield=56.79 g -38.5 g56.79 g× 100%Percent yield=32.21 %

Science

Chemistry

A student reacts benzene, C6H6, with bromine, Br2, to prepare bromobenzene, C6H5Br, and HBr. C6H6+ Br2 --> C6H5B + HBr What is the theoretical yield of bromobenzene in this reaction when 34.8 g of benzene reacts with 57.8 g of bromine? grams Which chemical is the limiting reactant? Select an answer If the actual yield of bromobenzene was 38.5 g, what was the percent yield? de

A student reacts benzene, C6H6, with bromine, Br2, to prepare bromobenzene, C6H5Br, and HBr. C6H6+ Br2 --> C6H5B + HBr What is the theoretical yield of bromobenzene in this reaction when 34.8 g of benzene reacts with 57.8 g of bromine? grams Which chemical is the limiting reactant? Select an answer If the actual yield of bromobenzene was 38.5 g, what was the percent yield? de

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

What is the theoretical yield (in grams) of acetic acid, HC₂H3O2, when 1.824 g of carbon dioxide reacts with 1.282 g of hydrogen according to the following balanced reaction: 2 CO2 (g) + 4H2(g) → HC₂H3O2(g) + 2 H₂O(g) Report your answer to 4 sig figs and do not include units in the answer.
For the following reaction, 4.80 grams of hydrogen gas are allowed to react with 9.37 grams of ethylene (C,H4). hydrogen (g) + ethylene (C¿H4) (g)- - ethane (C,Hg) (g) What is the maximum amount of ethane (C,Hô) that can be formed? | grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? | grams
For the following reaction, 10.3 grams of propane (C3Hg) are allowed to react with 21.3 grams of oxygen gas . propane (C3H8) ( g) + oxygen ( g )- - carbon dioxide ( g ) + water (g ) What is the maximum amount of carbon dioxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams
For the following reaction, 0.760 grams of hydrogen gas are allowed to react with 15.2 grams of ethylene (C,H). hydrogen (g) + ethylene (C2H4) (g)→ ethane (C¿Hg) (g) What is the maximum amount of ethane (C,H6) that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 8 more group attempts remaining
Methane burns in oxygen to form carbon dioxide and water according to the equation below. CH4 (g) + O2 (g) --> CO2 (g) + H2O (g) Balance the equation and answer the question below. If 3.25 moles of CH4 reacts, how many moles of H2O will form?
When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. 2A1(s) + 6HCI(aq) –→ 2AICI,(aq) + 3H,(g) If 10.0 moles of hydrochloric acid are reacted with excess aluminum, how many moles of aluminum chloride could be formed? 12.0molAICI3 30.0molAICI3 18.0molAICI3 3.33mol AICI3
When 11.0 mol of NO reacts, the actual yield of N2 is 8.63 mol. What is the percent yield? 4NH3(g) + 6NO(g) --> 5N2(g) + 6H2O(l)
Determine percent yield for the reaction. Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is:2Mg(s)+O2(g)→2MgO(s)2Mg(�)+O2(�)→2MgO(�) When 10.1 g of Mg are allowed to react with 10.5 g of O2, 11.6 g of MgO are collected.
Ag2S and NaNO3 react according to the following unbalanced chemical equation: Ag2S + NaNO3 → AgNO3 + Na2S Calculate the theoretical yield (in grams) of Na2S in the reaction between 0.441 g Ag2S and 0.858 g NaNO3.
For the following reaction, 25.3 grams of sulfur dioxide are allowed to react with 5.89 grams of water. sulfur dioxide (g) + water (1) → sulfurous acid (H₂SO3) (g) What is the maximum amount of sulfurous acid (H₂SO3) that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams
Nitric Oxide can be made from the oxidation of ammonia as shown below. A reaction between 8.5 g of NH3 and 25.0 g of O₂ produces 12.0 g of NO. What is the percent yield for this reaction? 4 NH3(g) + 5O₂(g) → 4 NO(g) + 6 H₂O(g) Molar masses (in g mol-¹): NH3 17.034, 0₂ 32.00, NO 30.01, H₂O 18.016
A 3.82-g sample of magnesium nitride is reacted with 7.73 g of water: Mg3N2 + 3H2O → 2NH3 + 3MgO The yield of MgO is 3.60 g. What is the percent yield in the reaction?