-0.651-36.46Before(mol)Change(mol)After (mol)0Calculate the percent yield for the reaction shown below byconstructing a BCA table, determining the maximum grams of productthat can be produced, and calculating the percent yield. Complete Parts1-3 before submitting your answer.0.3493NEXT >The reaction was started with 1.00 mol of MnO₂ (86.94 g/mol) and 95.0 g of HCI (MW36.458 g/mol). Set up the table below that represents 100% yield with the givenreaction conditions. Ignore the water side product.1.00Question 69 of 70-0.349MnO₂ + 4 HCI-2MnO₂+95.0MnCl₂ + Cl₂ + 2 H₂O24.0024 HCI-95.0-4.00MnCl₂+2.611.30CI₂-2.61-1.30SubmitRESET0.65136.46+ 01131Calculate the percent yield for the reaction shown below byconstructing a BCA table, determining the maximum grams of productthat can be produced, and calculating the percent yield. Complete Parts1-3 before submitting your answer.MnO₂ + 4 HCI → MnCl₂ + Cl₂ + 2 H₂ONEXT >< PREVBased on your table from the previous step, determine the maximum number ofgrams of Cl₂ (MW 70.90 g/mol) that can be prepared.massci₂Question 69 of 70185192.3=246.2g323.826.6SubmitRESET27.2+

Answered: Image /qna-images/answer/88974b76-0199-4310-b76e-1ec44bcfa123.jpg

Calculate the percent yield for the reaction shown below by constructing a BCA table, determining the maximum grams of product that can be produced, and calculating the percent yield. Complete Parts 1-3 before submitting your answer. MnO₂ + 4 HCI → MnCl₂ + Cl₂ + 2 H₂O 1 Before (mol) Change (mol) After (mol) 2 NEXT > The reaction was started with 1.00 mol of MnO₂ (86.94 g/mol) and 95.0 g of HCI (MW 36.458 g/mol). Set up the table below that represents 100% yield with the given reaction conditions. Ignore the water side product. MnO₂ 3 + 4 HCI MnCl₂ + CI₂

Calculate the percent yield for the reaction shown below by constructing a BCA table, determining the maximum grams of product that can be produced, and calculating the percent yield. Complete Parts 1-3 before submitting your answer. MnO₂ + 4 HCI → MnCl₂ + Cl₂ + 2 H₂O 1 Before (mol) Change (mol) After (mol) 2 NEXT > The reaction was started with 1.00 mol of MnO₂ (86.94 g/mol) and 95.0 g of HCI (MW 36.458 g/mol). Set up the table below that represents 100% yield with the given reaction conditions. Ignore the water side product. MnO₂ 3 + 4 HCI MnCl₂ + CI₂

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

Consider the reaction 4Fe(s) +302(g) --> 2FE203(s) If you started with and consumed 4 moles of O2, how many moles of Fe2O3 would you produce?
MISSED THIS? Watch KCV: Limiting Reactant, Theoretical Yield, and Percent Yield, IWE: Limiting Reactant and Theoretical Yield; Read Section 4.4. You can click on the Review link to access the section in your eText. Iron(II) sulfide reacts with hydrochloric acid according to the reaction FeS (s) + 2HCl(aq)→FeCl₂ (aq) + H₂S(s) A reaction mixture initially contains 0.212 mol FeS and 0.636 mol HCl. Part A Once the reaction has reached completion, what amount (in moles) of the excess reactant is left? Express the answer in moles to three significant figures. n = Submit V ΑΣΦ Request Answer ? mol
When 33.9 g of CH4 reacts completely with excess chlorine yielding 52.0 g of HCI, what is the percentage yield, according to CH418) + Cl2g) - CH3Clg) + HClig) 67.2 34.8 65.2 32.8 O 77.3
Which substance is the limiting reactant when 4.00mol of cobalt, Co, are reacted with 8.0mol of lead(II)chloride PbCl2 in the following equation? Co+PbCl2→CoCl2+Pb Choices are Co PbCl2 CoCl2 Pb None
Step by step
Consider the following unbalanced equation. How many molecules of hydrogen will be required to produce 1.60 mol of water?H2 + O2 H2O
Question Completion Status: QUESTION 2 Balancing refers to the number of substances put INTO a reaction has to equal the number of substance PRODUCED by the reaction. This means there must be an understanding of HOW MANY of something is put into a reaction. Match the following description with the HOW MANY X of each substance. Answers may be used once, more than once, or not at all. Example: 6 X3Cl means I have/need 18 X's Example: 10 HX means I have/need 10 X's Etc. - - - - - 3 X20 ✓ 2 Cax 4 XF2 5 AIX3 1 PBX4 ✓2 KX Click Save and Submit to save and submit. Click Save All Answers to save all answers. -4-531 84 A. 2X's B. 5 C. 15 X's D. 3 E. 4 X's F. 10 G. None of these H. 6X's 1.20 3000 103931 401818 / HOTEL hole-he artbl Save All Answers
Please answer these questions
Silver tarnishes in the presence of hydrogen sulfide according to the given balanced equation. What mass (g) of "tarnish" (Ag2S) is produced when 205.96 g of Ag react completely? Give your answer to 2 decimal places. 4 Ag + 2 H2S + O2 ==> 2 Ag2S + 2 H2O Molar Mass, g/mol Ag 107.9 H2S 34.08 O2 32.00 Ag2S 247.80 H2O 18.02
Can you help me solve this question with filling in the blanks from the numbers bellow. Please make it look clear because I send it before and I didn’t understand
A sample of 0.75 g of hydrogen gas was obtained by reacting 32.5 g hydrochloric acid with 18.0 g magnesium. Mg(s) + 2 HCI (aq) --> MgCl₂ (aq) + H₂(g) What is the percent yield for this reaction? 100% 83% 41% 50%
A student obtained 11.4 g of PbSO4 from the following reaction. The student added 15 g of Lead (II) nitrate to 15 g of Sodium sulfate. Pb(NO3)2 + Na2SO4 --> PbSO4 + 2NaNO3 What is the percent yield of the PbSO4 the student obtained from the reaction? % Make sure that you show your work for all questions, including units and labels being cancelled within the stoichiometry work itself and units and labels on your final answer. write it out as