# Gas, that was produced after burning 14.2g of methanol and ethanol mixture were let through lime water. 50g of precipitated out. What is the mass percent of methanol in the mixture?

Jan 12, 2017

WARNING! Long answer! The mass percent of methanol is 67 %.

#### Explanation:

The pertinent equations are

${M}_{r} : \textcolor{w h i t e}{l l} 32.04 \textcolor{w h i t e}{m m m m m l l} 44.01$
$\textcolor{w h i t e}{m m} \text{CH"_3"OH" + "2O"_2 → "CO"_2 + "2H"_2"O}$

${M}_{r} : \textcolor{w h i t e}{m} 46.07 \textcolor{w h i t e}{m m m m m m l l} 44.01$
$\textcolor{w h i t e}{m m} \text{2C"_2"H"_5"OH" + "7O"_2 → "4CO"_2 + "2H"_2"O}$

${M}_{r} : \textcolor{w h i t e}{m m m m m l l} 44.01 \textcolor{w h i t e}{m l l} 100.09$
$\textcolor{w h i t e}{m m} \text{Ca(OH)"_2 + "CO"_2 → "CaCO"_3 + "H"_2"O}$

Step 1. Calculate the moles of ${\text{CO}}_{2}$

${\text{Moles of CO"_2 = 50 color(red)(cancel(color(black)("g CaCO"_3))) × (1 color(red)(cancel(color(black)("mol CaCO"_3))))/(100.09 color(red)(cancel(color(black)("g CaCO"_3)))) × ("1 mol CO"_2)/(1 color(red)(cancel(color(black)("mol CaCO"_3)))) = "0.500 mol CO}}_{2}$

Step 2. Calculate the moles of ${\text{CO}}_{2}$ from methanol and from ethanol

Let mass of methanol be $x \textcolor{w h i t e}{l} g$.

Then the mass of ethanol is $\left(14.2 - x\right) \textcolor{w h i t e}{l} \text{g}$

From methanol:

${\text{Moles of CO"_2 = x color(red)(cancel(color(black)("g methanol"))) × (1color(red)(cancel(color(black)("mol methanol"))))/(32.04 color(red)(cancel(color(black)("g methanol"))))× ("1 mol CO"_2)/(1 color(red)(cancel(color(black)("mol methanol")))) = "0.031 21"xcolor(white)(l) "mol CO}}_{2}$

From ethanol:

${\text{Moles of CO"_2 = (14.2- x) color(red)(cancel(color(black)("g ethanol"))) × (1 color(red)(cancel(color(black)("mol ethanol"))))/(46.07 color(red)(cancel(color(black)("g ethanol")))) × ("4 mol CO"_2)/(2 color(red)(cancel(color(black)("mol ethanol")))) = "0.043 41"(14.2-x)color(white)(l) "mol CO"_2 = ("0.6165 - 0.043 41"x) color(white)(l)"mol CO}}_{2}$

Step 3. Equate moles of ${\text{CO}}_{2}$

"0.031 21"xcolor(white)(l) color(red)(cancel(color(black)("mol CO"_2))) + ("0.6165 - 0.043 41"x) color(red)(cancel(color(black)("mol CO"_2))) = 0.500 color(red)(cancel(color(black)("mol CO"_2)))

$\text{0.031 21"x + "0.6165 - 0.04341} x = 0.500$

$\text{0.0122} x = 0.1165$

$x = \frac{0.1165}{0.0122} = 9.55$

$\text{Mass of methanol = 9.55 g}$

Step 4. Calculate mass percent of methanol

"% by mass" = "mass of component"/"total mass" × 100 % = (9.55 color(red)(cancel(color(black)("g"))))/(14.2 color(red)(cancel(color(black)("g")))) × 100 % = 67 %

Note: The answer can have only 2 significant figures, because that is all you gave for the mass of ${\text{CO}}_{2}$.