#9 show all work 8:17 1Done1 of 216. What is the final molarity that would result if you diluted 180.0 mL of a 1.55 MHCl solution to a final volume of 500.0 mL?7. Determine the volume of 1.5 M KI solution required for the reaction of 10.00grams lead(II) nitrate.Pb(NO;)2+2KIPbl2+2KNO;8. How many grams of KNO3 are contained in 245 mL of 3.00 M KNO3 solution?9. If the combustion of 4.00 grams of magnesium results in the actual yield of 5.00grams of magnesium oxide, what is the percent yield of the reaction?2Mg(s) + O2(g)2MgO(s)I!!

Writing the balanced equation,2 Mg (s) + O2 (g) ------> 2 MgO (s) mass of Mg used in this…

Answered: 9. If the combustion of 4.00 grams of…

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...

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#9 show all work

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6. What is the final molarity that would result if you diluted 180.0 mL of a 1.55 M
HCl solution to a final volume of 500.0 mL?
7. Determine the volume of 1.5 M KI solution required for the reaction of 10.00
grams lead(II) nitrate.
Pb(NO;)2
+
2KI
Pbl2
+
2KNO;
8. How many grams of KNO3 are contained in 245 mL of 3.00 M KNO3 solution?
9. If the combustion of 4.00 grams of magnesium results in the actual yield of 5.00
grams of magnesium oxide, what is the percent yield of the reaction?
2Mg(s) + O2(g)
2MgO(s)
I!!

Expert Solution

Calculation of number of moles of MgO produced from 4 g of Mg

Writing the balanced equation,
2 Mg (s) + O₂ (g) ------> 2 MgO (s)

mass of Mg used in this reaction = 4 g
Molar mass of Mg = 24.305 g/mol
Moles of Mg = 4 g / 24.305 g/mol = 0.1645 mol

According to the reaction stoichiometry,
2 moles of Mg gives 2 moles of MgO
So, 0.1645 mol of Mg gives ________ moles of MgO
Cross multiply to find the moles of MgO produced from 0.1645 mol of Mg
(0.1645 mol Mg) x ( 2 mol MgO / 2 mol Mg) = 0.1645 mol MgO

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