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=
Stacey Warren - Expert brainly.com
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first write the balanced equation
i did but i am still confused
like i balanced it like this: C3H8+5o2-->3Co2+4h20
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find the no. of moles of C3H8
moles = mass divided by molar mass
what should i do after i get the moles
moles of CO2 = 3 times of moles of C3H8
use the same equation moles = mass/ molar mass
then we can find mass
should i find the moles of all of them first
now how do i find the mass
mass = molar mass * moles
the masses i got don't add up to 350 idk y
thank you though :)
wait I'll check for u
Mass of CO2 is 150 g approximately
Mass of H2O = 81.8 g approximately
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