The reaction below shows the combustion of butane (C4H10). Why is the mole ratio (8 moles CO2/2 mol C4H10) used to determine the amount of carbon dioxide (CO2) produced when butane (C4H10) combusts in excess oxygen?

C4H10 + O2 CO2 + H2O + heat

Butane has a mass that is 4 times greater than CO2, so only 2 moles of C4H10 are needed to produce 8 moles of CO2.

Carbon dioxide has a volume of about one-quarter that of C4H10, so 4 moles of CO2 are needed to react with each mole of C4H10.

During combustion, CO2 is approximately 4 times more reactive than C4H10, so less C4H10 will parti

Question

The reaction below shows the combustion of butane (C4H10). Why is the mole ratio (8 moles CO2/2 mol C4H10) used to determine the amount of carbon dioxide (CO2) produced when butane (C4H10) combusts in excess oxygen?

C4H10 + O2  CO2 + H2O + heat

 Butane has a mass that is 4 times greater than CO2, so only 2 moles of C4H10 are needed to produce 8 moles of CO2.

 Carbon dioxide has a volume of about one-quarter that of C4H10, so 4 moles of CO2 are needed to react with each mole of C4H10.

 During combustion, CO2 is approximately 4 times more reactive than C4H10, so less C4H10 will parti

lovelyharmonics · Answer

@Abmon98

lovelyharmonics · Answer

@hartnn

abmon98 · Answer

2C4H10+13O2-->8CO2+10H2O
C4H10:((12*4)+10)=58grams
CO2: 12+32=44 grams
1st Statment is incorrect

abmon98 · Answer

@lovelyharmonics are you sure that this question is complete??