C7H16 + 11 O2 ––> 7 CO2 + 8 H2O how many moles of carbon dioxide are produced if 5.0 mol of heptane...?

Question

how many moles of carbon dioxide are produced if 5.0 mol of heptane, C7H16 is burned.

Another chemistry question

please help and please explain how you got the answer

because I'm very confused =S

Thanks =)

Answer 1

it is very easy to do, when one of the moles of the reactant has been known, thus to calculate the other moles of the products or the other reactant just simply use the coefficient of the reaction.

because the moles of heptane is known so for calculate the moles of CO2 you use the ratio of the coefficient of the reactant of CO2 and C7H16, as:

C7H16 + 11 O2 ––> 7 CO2 + 8 H2O

moles CO2

= coeff CO2 / coeff C7H16 x moles of C7H16

= 7/1 x 5

= 35 moles

thus CO2 that produce from the reaction is 35 moles

so by this simple calculation you can calculate the O2 that needs for the reaction:

moles O2

= 11/1 x 5

= 55 moles

(if we use the coefficient ratio of moles O2 and H7H16)

or

moles O2

= 11/7 x 35

= 55 moles

((if we use the coefficient ratio of moles O2 and CO2)

but it should have the same results

I hope you get the point

Answer 2

C7H16 + 11 O2 ––> 7 CO2 + 8 H2O.

if 1mole of heptane is used 7 mole of CO2 is produced.

if we used 5 mole of heptane ––> ? moles of CO2.

1 mole of C7H16––> 7 mole of CO2.

5 mole of C7H16––> 5X7/1=35 MOLE OF CO2.

Answer 3

35 mols CO2 because:

5.0 mols C7H16 x 7 CO2/1 mol C7H16

For every 1 mol of C7H16 burned, 7 mols of CO2 are produced according to your equation.So using stoichiometry(fraction conversion)= 5 x (7/1)= 35