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

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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 HClag) reacts with 150. mL of 0.500 M KMNO4.
- 2 MnCl2(ag) + 5 Cl2(g) + 8 H2On+2 KCI@)
O 12.0 mol H20
O 0.300 mol H20
O 0.0375 mol H20
O 0.600 mol H20
O 0.0188 mol H20
Transcribed Image Text: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 H20 O 0.300 mol H20 O 0.0375 mol H20 O 0.600 mol H20 O 0.0188 mol H20
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