How many grams of CO2 is produced in the combustion of 100g butane?

Answer 1

C4H10 + 6.5 O2 -----> 4CO2 + 5H20

Mr of Butane = 58

5.8/58 = 0.1 So there is 0.1 moles of butane reacting

1 Mole = 24dm3 so 0.1 moles (5.8g of butane) = 0.1x24(dm3) = 2.4dm3

So then you see in the equation 4 moles of CO2 to 1 mole of C4H10

Thus 2.4 x 4 = Volume of CO2 = 9.6dm3

I think this is right, correct if wrong.

Answer 2

The reaction that takes place is the following:

2C4H10 + 13O2 = 8CO2 + 10H2O

If you have 100 grams of butane and butane's molecular weight is M(C4H10)=12*4+1*10=58 amu, then the molarity will be:

n=m/M=100/58=50/29 mol. (NB! That is per 2 molecules)

To get the molarity of only one molecule in the reaction, divide the n, you found above, by 2:

(50/29)/2=25/29

Next, to find the molarity of the CO2, simply multiply the 25/29 by 8 because that's how many molecules are produced in the reaction:

(25/29)*8=225/29 mol.

Now, final step isto calculate the weight of the CO2 gas emitted in the reaction:

M(CO2)=44 amu

m=nM=(225/29)*44 ~ 341,379 grams.

FINAL ANSWER: When burning 100 grams of butane, about 341,379 grams of cabon dioxide will be produced.

Answer 3

When butane (C4H10) is combusted, it reacts with oxygen to produce carbon dioxide (CO2) and water. The balanced chemical equation for this reaction is 2C4H10 + 13O2 -> 8CO2 + 10H2O.

From the equation, we can see that 2 moles of butane produce 8 moles of carbon dioxide. The molar mass of butane is approximately 58.12 g/mol, and the molar mass of carbon dioxide is approximately 44.01 g/mol.

Therefore, to find the amount of CO2 produced from 100g of butane, we calculate as follows:

100g butane * (8 mol CO2 / 2 mol butane) * (44.01 g CO2 / 1 mol CO2) = 176.04 g CO2.

So, 176.04 grams of CO2 would be produced when 100g of butane is combusted.