Question

50 grams of propane (c3h8) is oxidized (combusted/burned) completely to co2 and h20 by reaction with 300 grams of pure oxygen (o2). What is the composition of the reaction products in grams?
grams of c02=
grams of h20=
grams of c3h8=
grams of 02=

Answer 1

first write the balanced equation

Answer 2

i did but i am still confused
like i balanced it like this: C3H8+5o2-->3Co2+4h20

Answer 3

yep

Answer 4

find the no. of moles of C3H8
moles = mass divided by molar mass

Answer 5

what should i do after i get the moles

Answer 6

moles of CO2 = 3 times of moles of C3H8

Answer 7

use the same equation moles = mass/ molar mass

Answer 8

then we can find mass

Answer 9

of CO2

Answer 10

should i find the moles of all of them first

Answer 11

yeah yeah

Answer 12

oh ok

Answer 13

now how do i find the mass

Answer 14

mass = molar mass * moles

Answer 15

the masses i got don't add up to 350 idk y

Answer 16

thank you though :)

Answer 17

wait I'll check for u

Answer 18

Mass of CO2 is 150 g approximately

Answer 19

Mass of H2O = 81.8 g approximately

Answer 20

Looking at the stoichiometric ratio we can see the ratio of moles of C3H8: O2 is 1:5
That means 1 moles of C3H8 reacts with 5 moles of O2
We know the no. of moles of C3H8 that was 1.136 right?
That means to oxidise C3H8 completely we need (1.136 *5) moles right?
so no. of moles of O2 needed = 5.68 moles
But we are said that we have 300 g of O2
Using that mass we should find the no. of moles of O2 . The total O2 present
Total moles = 300/32 = 9.375
But to complete the reaction we only need 5.68 moles of O2
That means O2 is in excess. So that will remain in the products side.
The remaining would be 9.375 - 5.68 = 3.695 moles
Now we can find the mass of O2 remaining
mass = 3.695 * 32= 118.24g

Answer 21

oh okay thank you so much

Answer 22

No problem ... hope you got it :)