How many moles of glycerin (c3h5(oh)3) are consumed in this reaction? 14kmno4 + 4c3h5(oh)3 es001-1.jpg 7k2co3 + 7mn2o3 + 5co2 + 16h2o

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

2.48  0.709

i got it right on edg so enjoy

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"