Combustion of mixture ethanol and methanol

Mar 2020
25
0
.
Now you are doing it again.

So I don't know if you have understood any of what I have written.
Yes. The amount of ethanol is 0.04 moles and the mass of methanol is 15% of 2.16 g
 
Apr 2015
133
94
Well I would agree with the 0.04 moles of ethanol, but I made the weight% of methanol as 17%, though that could be my famously sloppy arithmetic.

OK so the simultaneous equations are

a + b = 2.16

and


\(\displaystyle \frac{{96}}{{46}}a + \frac{{96}}{{64}}b = 4.32\)

Chemists usually prefer the simpler longer method with one unknown that I also showed you.

Obviously both ways will (should) reach the same answers.
 
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Mar 2020
25
0
.
Well I would agree with the 0.04 moles of ethanol, but I made the weight% of methanol as 17%, though that could be my famously sloppy arithmetic.

OK so the simultaneous equations are

a + b = 2.16

and


\(\displaystyle \frac{{96}}{{46}}a + \frac{{96}}{{64}}b = 4.32\)

Chemists usually prefer the simpler longer method with one unknown that I also showed you.

Obviously both ways will (should) reach the same answers.
Why do i keep getting a = 2.16 g?
 
Apr 2015
133
94
\(\displaystyle \frac{{96}}{{46}}a + \frac{{96}}{{64}}(2.16 - a) = 4.32\)

\(\displaystyle \frac{{48}}{{23}}a + \frac{{48}}{{32}}(2.16 - a) = 4.32\)

\(\displaystyle 32a + 23(2.16 - a) = \frac{{4.32*23*32}}{{48}}\)

\(\displaystyle 32a + \left( {23*2.16} \right) - 23a = 66.24\)

\(\displaystyle 32a - 23a = 66.24 - 50.14\)

\(\displaystyle a = \frac{{16.1}}{9} = 1.789g\)

\(\displaystyle \frac{{1.79g}}{{46}} = 0.0389mole\)

\(\displaystyle b = 2.16 - 1.79 = 0.37g\)

\(\displaystyle b\% = \frac{{0.37}}{{2.16}}*100 = 17.13\% \)
 
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