0.220 what?
....burned in an excess of .....and yields 0.343 etc.
0.343 what?
I've tried to solve this question multiple ways I just can't seem to get it. I just don't understand what I should be doing.
An organic liquid is a mixture of methyl alcohol () and ethyl alcohol (). A 0.220- sample of the liquid is burned in an excess of and yields 0.343 (carbon dioxide).
Set up two algebraic equations, one expressing the mass of carbon dioxide produced in terms of each reagent and the other expressing the mass of sample burned in terms of each reagent.
What is the mass of methyl alcohol () in the sample?
2 answers
If I assume 0.220 is grams.
If I assume burned in ..... is burned in excess of oxygen.
If I assume 0.343 is grams. Here is what you do.
Look at the balanced equations.
2CH3OH + 3O2 ==> 2CO2 + 4H2O
C2H5OH + 3O2 ==> 2CO2 + 3H2O
---------------------------
let x = grams methyl alcohol.
let y = grams ethyl alcohol.
-------------------------
In general this is what we do and I'll get to the equations in terms of x and y next.
---------------------
grams methanol + grams ethanol=0.220g.
grams methanol converted to g CO2 + grams ethanol converted to g CO2 = 0.343
now we put those last two into equation form.;
-------------------------------
x + y = 0.220
(44x/32) + (88y/46) = 0.343
Two equations in two unknowns. Solve simultaneously for x and y.
x is grams methanol.
-------------------------
Note: Where does the second equation come from?
44x/32 is this
x*(1 mol MeOH/molar mass MeOH)*(1 mo CO2/1 mol MeOH)*(44 g CO2/1 mol CO2) = 0.220.
88y/46 is this.
y*(1 mol EtOH/molar mass EtOH)*(2 mol CO2/1 mol EtOH)*(44 g CO2/ mol CO2) = 0.343
If I assume burned in ..... is burned in excess of oxygen.
If I assume 0.343 is grams. Here is what you do.
Look at the balanced equations.
2CH3OH + 3O2 ==> 2CO2 + 4H2O
C2H5OH + 3O2 ==> 2CO2 + 3H2O
---------------------------
let x = grams methyl alcohol.
let y = grams ethyl alcohol.
-------------------------
In general this is what we do and I'll get to the equations in terms of x and y next.
---------------------
grams methanol + grams ethanol=0.220g.
grams methanol converted to g CO2 + grams ethanol converted to g CO2 = 0.343
now we put those last two into equation form.;
-------------------------------
x + y = 0.220
(44x/32) + (88y/46) = 0.343
Two equations in two unknowns. Solve simultaneously for x and y.
x is grams methanol.
-------------------------
Note: Where does the second equation come from?
44x/32 is this
x*(1 mol MeOH/molar mass MeOH)*(1 mo CO2/1 mol MeOH)*(44 g CO2/1 mol CO2) = 0.220.
88y/46 is this.
y*(1 mol EtOH/molar mass EtOH)*(2 mol CO2/1 mol EtOH)*(44 g CO2/ mol CO2) = 0.343
1 answer