# Combustion of mixture ethanol and methanol

#### Helly

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

#### studiot

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.

Helly

#### Helly

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?

#### studiot

Why don't you post your working?

topsquark

#### Helly

Why don't you post your working?
$2a + 1.5(2.16-a) =4.32$
$0.5a=1.08$
$a=2.16$
96/46 = 2.08 i count it as 2, apparently it makes whole wrong

#### studiot

$$\displaystyle \frac{{96}}{{46}}a + \frac{{96}}{{64}}(2.16 - a) = 4.32$$

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

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

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

$$\displaystyle 32a - 23a = 66.24 - 50.14$$

$$\displaystyle a = \frac{{16.1}}{9} = 1.789g$$

$$\displaystyle \frac{{1.79g}}{{46}} = 0.0389mole$$

$$\displaystyle b = 2.16 - 1.79 = 0.37g$$

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

Last edited:
topsquark and Helly