# Combustion of mixture ethanol and methanol

Last edited:

#### studiot

I have had other things to do for the past couple of days so we never finished your last question.

It is important you answer questions here and attempt the part suggested, which you did not do.

The only purpose of our input is to help you so if you do not cooperate little progress will be made.

I would start this question as follows

You are asked both the mass and number of moles of ethanol so you can work on either weight or numbers of moles.

Choosing numbers of moles

Let a be the number of grammes of ethanol and b be the number of grammes of methanol we have 2 unknowns.

So we require 2 equations connecting these to solve this question.

Which gives us one equation

a + b = 2.16 (grammes)

So we need one more equation

C2H5OH + 3O2 = 2CO2 + 3H2O
CH3OH + 3/2 O2 = CO2 + 2H2O
If you combine these (add them up) you will get a second equation.
(I suggest you double the second one to get rid of the fraction)

Then substitute the molecular masses of each substance to get a mass equation in grammes.
Finally multiply through by a suitable number to make the mass of oxygen equal to that stated (4.32g)

You will then have a second equation in a and b

Can you solve a pair of simultaneous equations?

• Helly and topsquark

#### studiot

Since you won't answer a simple question I take it that you don't want help to be able to do these questions yourself.
you just want someone to do your work for you.

Thank you for wasting my time.

#### Helly

What should i do after 2CH3OH + C2H5OH + 6O2 = 4CO2 + 7H2O?

#### studiot

Can you solve a pair of simultaneous equations?

#### Helly

Can you solve a pair of simultaneous equations?
The truth is, i dont even get what is pair of simultaneous equations? Its like multiple equations?

#### studiot

The truth is, i dont even get what is pair of simultaneous equations? Its like multiple equations?
OK that is fine..

So we can do this another way that is simpler but longer, that does not use simultaneous equations directly.

So let x be the weight of ethanol.

Then the weight of methanol is (2.16 - x) grammes.

Using the molecular weights (masses) and the equations of reaction

$$\displaystyle {C_2}{H_5}OH + 3{O_2} \to 2C{O_2} + 3{H_2}O$$

46g of ethanol require 96 grammes of oxygen for combustion
1g of ethanol requires 96/46 grammes of oxygen for combustion
xg of ethanol require x*(96/46) grammes of oxygen for combustion

$$\displaystyle 2C{H_3}OH + 3{O_2} \to 2C{O_2} + 4{H_2}O$$

64g of methanol require 96 grammes of oxygen for combustion
1g of methanol requires 96/64 grammes of oxygen for combustion
xg of methanol require (2.16 - x)*(96/64) grammes of oxygen for combustion

But this equals 4.32 grammes

That is

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

Can you work out the next part?

• Helly

#### studiot

xg of methanol require (2.16 - x)*(96/64) grammes of oxygen for combustion

(2.16 - x)g of methanol require (2.16 - x)*(96/64) grammes of oxygen for combustion

• Helly

#### Helly

OK that is fine..

So we can do this another way that is simpler but longer, that does not use simultaneous equations directly.

So let x be the weight of ethanol.

Then the weight of methanol is (2.16 - x) grammes.

Using the molecular weights (masses) and the equations of reaction

$$\displaystyle {C_2}{H_5}OH + 3{O_2} \to 2C{O_2} + 3{H_2}O$$

46g of ethanol require 96 grammes of oxygen for combustion
1g of ethanol requires 96/46 grammes of oxygen for combustion
xg of ethanol require x*(96/46) grammes of oxygen for combustion

$$\displaystyle 2C{H_3}OH + 3{O_2} \to 2C{O_2} + 4{H_2}O$$

64g of methanol require 96 grammes of oxygen for combustion
1g of methanol requires 96/64 grammes of oxygen for combustion
xg of methanol require (2.16 - x)*(96/64) grammes of oxygen for combustion

But this equals 4.32 grammes

That is

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

Can you work out the next part?
• 