C60H87O23N12P + ? O2
to get ? HNO3 + ? CO2 + ?H2O + ?H3PO4
start with C. Put 60 CO2
HNO3 + 60CO2 + ?H2O + ?H3PO4
Now look at N Put a 12 on the nitric acid. 12HNO3 + 60CO2 + ?H2O + ?H3PO4
now keep the P at one, so we work at on the H on water.
we have now 12+1+?=87 or 74H2O
12HNO3 + 60CO2 + 74H2O + H3PO4
finally, check O's. On the right, we have36+120+74+4=234
on the left we have 23+? which=234
or we have 211 O or 211/2 O2. Lessor minds don't like to balance with fractions, so we double all else.
FInal 2C60H87O23N12P+ 211O2>>24HNO3 + 120CO2 + 144H2O + 2H3PO4
go through that on paper and double check, it is easy to err when doing this on a keyboard.
Long question but its my last one for homework and im stuck on this word problem.
Biomass is represented by the empirical formula C60H87O23N12P. Write a balance equation for the oxidation of C60H87O23N12P with O2 to form CO2, H2O, HNO3, and H3PO4 beginning with 1000g of C60H87O23N12P.
After oxidation how many g of CO2 will be released in the air?
3 answers
2C60H87O23N12P +173O2 ==>120CO2 + 72H2O + 24HNO3 + 2H3PO4
Check this carefully.
mols biomass = grams/molar mass
Using the coefficients in the balanced equation, convert mols biomass to mols CO2.
Now convert mols CO2 to grams. g = mols CO2 x molar mass CO2
Check this carefully.
mols biomass = grams/molar mass
Using the coefficients in the balanced equation, convert mols biomass to mols CO2.
Now convert mols CO2 to grams. g = mols CO2 x molar mass CO2
I've gone cross eyed with these large numbers but I don't believe H and O balance in your equation.