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?
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.
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
I've gone cross eyed with these large numbers but I don't believe H and O balance in your equation.
To find out how many grams of CO2 will be released in the air after the oxidation of C60H87O23N12P, we need to calculate the molar masses and stoichiometry of the reaction.
Step 1: Calculate the molar masses:
The molar mass of C60H87O23N12P can be calculated by adding up the individual molar masses of the elements in the empirical formula. The molar masses of the elements are as follows:
- C (Carbon): 12.01 g/mol
- H (Hydrogen): 1.01 g/mol
- O (Oxygen): 16.00 g/mol
- N (Nitrogen): 14.01 g/mol
- P (Phosphorus): 30.97 g/mol
Now we can calculate the molar mass of C60H87O23N12P:
Molar mass = (60 * 12.01 g/mol) + (87 * 1.01 g/mol) + (23 * 16.00 g/mol) + (12 * 14.01 g/mol) + (1 * 30.97 g/mol)
Step 2: Calculate the moles of C60H87O23N12P:
To find the number of moles of C60H87O23N12P, we divide the given mass (1000g) by its molar mass calculated in Step 1.
Moles of C60H87O23N12P = Mass of C60H87O23N12P / Molar mass of C60H87O23N12P
Step 3: Stoichiometry of the reaction:
The balanced equation for the oxidation of C60H87O23N12P with O2 is not provided, so without knowing the exact reaction, it is not possible to determine the stoichiometric coefficients for CO2, H2O, HNO3, and H3PO4 directly. However, we can assume that the coefficients are whole numbers and proceed with the calculation based on that assumption.
For example, if we assume that the balanced equation is:
C60H87O23N12P + xO2 → yCO2 + zH2O + aHNO3 + bH3PO4
We need to find the ratio of the coefficients of C60H87O23N12P to CO2. Assuming that y = 1, we can calculate the value of x using the balanced equation and the mole ratio.
Moles of C60H87O23N12P / Moles of CO2 = Coefficient ratio (x/y)
Moles of CO2 = (Moles of C60H87O23N12P / Moles of CO2) * Coefficient ratio
Step 4: Convert moles of CO2 to grams:
To convert moles of CO2 to grams, we multiply the number of moles by the molar mass of CO2, which is 44.01 g/mol.
Grams of CO2 = Moles of CO2 * Molar mass of CO2
It is important to note that without the exact balanced equation for the reaction, we can only make assumptions and calculate using those assumptions. To get an accurate answer, the balanced equation must be provided.
I hope this explanation helps you understand how to approach the problem. Please let me know if you need any further assistance!