Organic Chemistry II I can\'t figure out these two problems Balance the equation

Question

Organic Chemistry II I can't figure out these two problems

Balance the equation for the reaction for the oxidation of isoborneol to camphor using bleach by filling in the stoiciometric coefficients: 1 C_10 H_18 O(isoborneol)+1 NaOCl right arrow 1 C_10 H_16 O + 1 H_2 O+1 NaCl Determine the limiting reactant: isoborneol Determine the theoretical yield: {2:NM=25:0.1} mmol which is grams. If 3.358 grams of product are isolated, the percent yield would be percent The literature melting point of camphor is 175 degree Celsius. If a sample of camphor that was prepared by oxidation of isoborneol in aqueous solution has a melting point of 101 degree Celsius. determine the percent impurity by mass if: The impurity is water: percent by mass. The impurity is isoborneol: percent by mass.

Explanation / Answer

Molar mass of Isoborneol = 154 g/mol

Molar mass of camphor = 152 g/mol

Given that moles of isoborneol = 25 mmol

mass of isoborneol = 0.025 mol x 154 g/mol = 3.85 g

Given reaction is

Isoborneol + NaOCl    -------------> camphor

1 mol                                        1 mol

154 g                                         152 g

3.85g                                          ?           = (3.85/154) x 152 g = 3.8 g of camphor

This is theoretical yield of camphor.

Theoretical yield of camphor = 3.8 g

Given that actual yield of camphor = 3.358 g

Therefore,

% yield of camphor = (actual yield/ theoretical yield) x 100

                              = (3.358 / 3.8) x 100

                             = 88.3 %

                                         

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