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How many moles of glycerin (C3H5(OH)3) are consumed in this reaction?

14KMnO4 + 4C3H5(OH)3 es001-1.jpg 7K2CO3 + 7Mn2O3 + 5CO2 + 16H2O

How many moles of glycerin (C3H5(OH)3) are consumed in this reaction? 14KMnO4 + 4C-example-1

2 Answers

6 votes

Answer:

2.48 0.709

i got it right on edg so enjoy

User Jchrbrt
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2 votes

Ans: 0.709 moles glycerin

The given reaction is:

14KMnO4 + 4C3H5(OH)3 → 7K2CO3 + 7Mn2O3 + 5CO2 + 16H2O

a) Based on the reaction stoichiometry:

5 moles of CO2 requires 14 moles of KMnO4

Therefore, 0.886 moles of CO2 would correspond to:

= 0.886 moles CO2 * 14 moles KMnO4/5 moles CO2

= 2.48 moles

b) Again from the reaction stoichiometry:

5 moles of CO2 requires 4 moles of glycerin

therefore, 0.886 moles of CO2 would consume:

= 0.886 moles CO2 * 4 moles glycerin/5 moles CO2

= 0.709 moles

User Anthony Geoghegan
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