Question

2.30 g of ethanol were mixed with an excess of aqueous acidified potassium

dichromate(VI). The reaction mixture was then boiled under reflux for one hour. The

desired organic product was then collected by distillation. The yield of product was

60.0 %. What mass of product was collected?

a. 1.32 g b. 1.38 g c. 1.80 g d. 3.20 g​

Answer 1

Answer:

Option C is correct

Explanation:

We know that we will need molar masses and moles, so let's gather all the information in one place

Step 1. Write the chemical equation

The product of the reaction is acetic acid, (CH3COOH).

       46.07                                            60.05

3CH3CH2OH+2Cr2O2-7+16H+→3CH3COOH+4Cr3++11H2O

Mass/g:mml2.30

Moles:mmml0.049 92

Step 2. Calculate the moles of ethanol

Moles of ethanol=2.30g /46.07g ethanol

=0.049 92 mol ethanol

Step 3. Calculate the moles of acetic acid

Moles of acetic acid=0.049 92g ethanol×1 mol acetic acid / 1mol ethanol

=0.049 92 mol acetic acid

Step 4. Calculate the theoretical yield of acetic acid

Theor. yield=0.049 92mol acetic acid×60.05 g acetic acid /1mol acetic acid

=2.998 g acetic acid

Step 5. Calculate the actual yield

Actual yield = 2.998 x 60/100

=1.79 g