in outo engine whith no pollusion controls, about 5% of the fuel(assume 100%actane c8H18)is unburned .calculate the mass and volume ratio of Co to C8H18 in exhaust gas?
2C8H18+17O2 =16CO+18H2O
To calculate the mass and volume ratio of carbon monoxide (CO) to octane (C8H18) in the exhaust gas, we need to follow these steps:
1. Calculate the molar mass of C8H18:
- Carbon (C) has a molar mass of 12.01 g/mol, and hydrogen (H) has a molar mass of 1.01 g/mol.
- Multiply the atomic masses by their respective subscripts and sum them up:
Molar mass of C8H18 = (12.01 g/mol * 8) + (1.01 g/mol * 18) = 114.23 g/mol
2. Calculate the molar mass of CO:
- Carbon (C) has a molar mass of 12.01 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol.
- Molar mass of CO = (12.01 g/mol) + (16.00 g/mol) = 28.01 g/mol
3. Determine the number of moles of C8H18:
- Assuming 100% of the fuel is C8H18 and 5% is unburned, we can calculate the moles of C8H18 that is unburned.
- Let's assume we have 100 moles of C8H18 in the fuel.
- Unburned moles of C8H18 = 5/100 * 100 moles = 5 moles
4. Calculate the moles of CO produced:
- According to the balanced equation, 2 moles of C8H18 produce 16 moles of CO.
- Moles of CO produced = 5 moles of C8H18 * (16 moles of CO / 2 moles of C8H18) = 40 moles
5. Calculate the mass of CO produced:
- Mass of CO = Moles of CO * Molar mass of CO
- Mass of CO = 40 moles * 28.01 g/mol = 1120.4 g
6. Calculate the volume ratio of CO to C8H18:
- Since the number of moles is directly proportional to the volume under constant temperature and pressure, we can assume the ratio of moles as the ratio of volumes.
- The volume ratio of CO to C8H18 = 40 moles CO / 5 moles C8H18 = 8
Therefore, the mass ratio of CO to C8H18 in the exhaust gas is 1120.4 g : 114.23 g, and the volume ratio is 8:1.