Write the balanced equation for the reaction that occurs when methanol, cheoh () is burned in air. what is the coefficient of oxygen in the balanced equation? a) 1 b) 2 c) 3 d) 4 e) 3/2

The balanced equation for methanol when is burned in the air, is

CH3OH + O2 >  3/2 CO2  +2 H2O and as you see coefficient of oxygen is 3/2

Explanation:

When you always burn something you are doing combustion. The reactives are your compound + O2, and as products you have CO2 and H2O

The coefficient of oxygen is 3 ( option c)

Explanation:

Methanol = CH3OH

The process of burning = adding oxygen (O2)

In this process methanol will be consumed and CO2 and H2O will be produced.

CH3OH + O2 → CO2 + H20

First we'll balance on both sides all H-atoms: left we have 4x H and on the right side we have 2x H so we have to multiply the right side ( H2O) by 2.

This gives us:   CH3OH + 02 → CO2 + 2 H2O

Since we have now, 3x O on the left side and 4x O on the right side, it looks like we don't have to multiply H2O on the right side by 2 but by 4, and then we multiply CH3OH on the left side by 2

This gives us: 2 CH3OH + O2 → CO2 + 4H2O

Now we have on both side 8x H

On the left side we have 2x C, and 1x C on the right side. So we have to multiply CO2 on the right side by 2

This gives us: 2 CH3OH + O2 → 2 CO2 + 4H20

Now we have on both side 8x H and 2x C

On the left side we have 4x O but on the right side we have 8x O

We have to multiply O2 on the left side by 3, to equal this.

So finally, we'll have

2 CH3OH + 3O2 → 2 CO2 + 4H2O

This shows us that the coefficient of oxygen is 3 ( option c)

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o

lab safety rules

explanation: