# 2.30 g of ethanol were mixed with aqueous acidified potassium dichromate(VI). The desired product was collected by immediate distillation under gentle warming. The yield of product was 70.0%. What mass of product was collected?

May 22, 2017

You obtained 2.10 g of product.

#### 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, ($\text{CH"_3"COOH}$).

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m m m} 46.07 \textcolor{w h i t e}{m m m m m m m m m m m m m m m} 60.05$
$\textcolor{w h i t e}{m m m m} \text{3CH"_3"CH"_2"OH" + "2Cr"_2"O"_7^"2-" + "16H"^"+" → "3CH"_3"COOH" + "4Cr"^"3+" + "11H"_2"O}$
$\text{Mass/g} : \textcolor{w h i t e}{m m l} 2.30$
$\text{Moles:"color(white)(mmml)"0.049 92}$

Step 2. Calculate the moles of ethanol

$\text{Moles of ethanol" = 2.30 color(red)(cancel(color(black)("g ethanol"))) × "1 mol ethanol"/(46.07 color(red)(cancel(color(black)("g ethanol")))) = "0.049 92 mol ethanol}$

Step 3. Calculate the moles of acetic acid

$\text{Moles of acetic acid" = "0.049 92" color(red)(cancel(color(black)("g ethanol"))) × "1 mol acetic acid"/(1 color(red)(cancel(color(black)("mol ethanol")))) = "0.049 92 mol acetic acid}$

Step 4. Calculate the theoretical yield of acetic acid

$\text{Theor. yield" = "0.049 92" color(red)(cancel(color(black)("mol acetic acid"))) × "60.05 g acetic acid"/(1 color(red)(cancel(color(black)("mol acetic acid")))) ="2.998 g acetic acid}$

Step 5. Calculate the actual yield

$\text{Actual yield" = 2.998 color(red)(cancel(color(black)("g theor."))) × "70.0 g"/{100 color(red)(cancel(color(black)("g theor.")))) = "2.10 g}$