Use the following balanced equation to answer the solution stoichiometry problem.2 KMNO4(aq) + 16 HCl(aq)Calculate the moles of H20 produced if excess HClag) reacts with 150. mL of 0.500 M KMNO4.- 2 MnCl2(ag) + 5 Cl2(g) + 8 H2On+2 KCI@)O 12.0 mol H20O 0.300 mol H20O 0.0375 mol H20O 0.600 mol H20O 0.0188 mol H20

The number of mol (N) of a compound present in a solution with molarity equals to C (M) and a volume of V mL is calculated by using the following expression: N=C×V1000 mol ................... (1)The number of mol of KMnO4 (NKMnO4) present in 150. mL of 0.500 M KMnO4 is calculated by using equation 1 as shown below: NKMnO4=0.500×150.1000 mol =0.0750 mol As HCl is present in excess amount, therefore, the number of mol of H2O produced depends on the number of mol of KMnO4 reacted.According to the balanced equation- 2 mol of KMnO4 produces 8 mol of H2O So, 0.0750 mol of KMnO4 produces (82×0.0750) mol or 0.300 mol of H2O. 0.300 mol of H2O produced.

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Use the following balanced equation to answer the solution stoichiometry problem. 2 KMNO4(aq) + 16 HCl(aq) Calculate the moles of H20 produced if excess HCI(ag) reacts with 150. mL of 0.500 M KMnO4. 2 MnCl2(aq) + 5 Cl2) + 8 H2O+2 KCI6) O 12.0 mol H2O O 0.300 mol H20 0.0375 mol H20 0.600 mol H2O 0.0188 mol H2O

Use the following balanced equation to answer the solution stoichiometry problem. 2 KMNO4(aq) + 16 HCl(aq) Calculate the moles of H20 produced if excess HCI(ag) reacts with 150. mL of 0.500 M KMnO4. 2 MnCl2(aq) + 5 Cl2) + 8 H2O+2 KCI6) O 12.0 mol H2O O 0.300 mol H20 0.0375 mol H20 0.600 mol H2O 0.0188 mol H2O

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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