well i really need help with this question, but i really want to get this stoichiometry prob. to do 1 myself!!?
biodiesel makes use of plants ability to fix atmoshpere carbon by photosynthesis. Many companies and individuals are now using biodiesel as a fuel in order to reduce thir carbon footprint. Biodiesel can by synthesized from from vegetable oil according to the following reaction.
C6H5O6R3(vegetable oil)+ 3CH3OH(methanol)—> C3H5(OH)3(glycerol)+CH3OCOR(biodiesel)
a) a student reacted a pure sample of vegetable oil ( where R= C17H33) with methanol.
the raw data for the reaction is below.
mass of oil=1010.0 g
mass of methanol=200.0 g
mass of sodium hydrozide=3.5 g
mass of biodiesel produced=811.0 g
the relative molecular mass of the oil used by the student is 885.5. calculate the amount in moles of the oil and the methanol used, and hence the amnount in moles of the excess methanol used.
b) calculate the percentage yield of biodiesel obtained in this process.
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i expect you’ll want to balance your reaction; you have an R3 on lhs, so you need 3R on rhs to balance.
a) 1 mol pure vegetable oil = 885.5g
[check: add up the total numbers of each type of atom, times its mass number =
6C +3x(17C) = 57C; 57 x 12.0 = 684
5H +3x(33H) = 104H; 104 x 1.0 = 104
6O; 6 x 16.0 = 96
684+104+96 = 884 g/mol; this agrees with your 885.5 g/mol once we boost the precision and account for likely differences in isotopic composition ]
1010.0g / 885.5 g/mol = 1.1406 mol
methanol: 3C=3×12.0=36.0; 10H=10.0; 1O=16.0; 36.0+10.0+16.0= 62.0 g/mol
200g / 62.0 g/mol = 3.2258 mol
so you have 1.1406 mol oil reacting with 3.23 mol methanol.. but the reaction wants ratio of 1:3, so you have excess vegetable oil.
1.1406 - 3.2258/3 = 0.065 mol excess vegetable oil.
b) yield is 811.0 g of CH3OCOR =
2C+3H+2O+17C+33H = 19C+36H+2O = 19×12.0+36.0+2×16.0= 296 g/mol;
811.0g / 296 g/mol = 2.74 mol
1.14 mol vegetable oil would produce 3×1.14 = 3.42 mol biodiesel at 100% yield. but you obtained 2.74 mol. 2.74/3.42 = 0.801, so 80.1% yield.
you might increase the yield by adding excess methanol.