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?

1 Answer
Jan 12, 2017

Answer:

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

Explanation:

The pertinent equations are

#M_r:color(white)(ll)32.04color(white)(mmmmmll)44.01#
#color(white)(mm)"CH"_3"OH" + "2O"_2 → "CO"_2 + "2H"_2"O"#

#M_r:color(white)(m)46.07color(white)(mmmmmmll)44.01#
#color(white)(mm)"2C"_2"H"_5"OH" + "7O"_2 → "4CO"_2 + "2H"_2"O"#

#M_r:color(white)(mmmmmll)44.01color(white)(mll)100.09#
#color(white)(mm)"Ca(OH)"_2 + "CO"_2 → "CaCO"_3 + "H"_2"O"#

Step 1. Calculate the moles of #"CO"_2#

#"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 #"CO"_2# from methanol and from ethanol

Let mass of methanol be #xcolor(white)(l) g#.

Then the mass of ethanol is #(14.2 - x) color(white)(l)"g"#

From methanol:

#"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:

#"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 #"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)))#

#"0.031 21"x + "0.6165 - 0.04341"x = 0.500#

#"0.0122"x = 0.1165#

#x = 0.1165/0.0122 = 9.55#

#"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 #"CO"_2#.