i.Calculate the volume of carbon ivoxide given off at stp when 20gram of octane is burnt in excess air (c=12,1+=1 gmv of any gass is 22.4dm^3 at stp). ii. Deduce the formula of a hydrocarbon with general formula cnH2n-2 of its molar mass in 110.

C8H16 + 12O2 ==> 8CO2 + 8H2O

mols C8H18 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols C8H18 to mols CO2.
Then mols CO2 x 22.4 dm^3/mol = ?dm^3 CO2

ii.
CnH2n-2 has an empirical mass of
12n + 2n - 2 = 110
14n = 108
n = about 7.7 which rounds to 8 so
C8H(2*8-2) = C8H14
Check: 12*8 = 96
14*1 = 14
96+14 = 110.

To calculate the volume of carbon dioxide (CO2) given off when 20 grams of octane (C8H18) is burned in excess air at STP (Standard Temperature and Pressure), we need to follow these steps:

Step 1: Convert the mass of octane to the number of moles.
The molar mass of octane (C8H18) can be calculated by adding the atomic masses of carbon (C) and hydrogen (H).

Molar mass of C8H18 = (8 * atomic mass of carbon) + (18 * atomic mass of hydrogen)
= (8 * 12.01 g/mol) + (18 * 1.01 g/mol)
= 96.08 g/mol + 18.18 g/mol
= 114.26 g/mol

Number of moles of octane = mass of octane / molar mass
= 20 g / 114.26 g/mol
= 0.175 moles

Step 2: Determine the stoichiometry of the reaction.
The balanced equation for the combustion of octane is:
2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O

According to the equation, every 2 moles of octane produces 16 moles of CO2.
So, the number of moles of CO2 produced can be calculated as follows:

Number of moles of CO2 = (0.175 moles of octane) * (16 moles of CO2 / 2 moles of octane)
= 1.4 moles

Step 3: Calculate the volume of CO2 at STP.
According to Avogadro's law, 1 mole of any gas occupies 22.4 dm^3 (liters) at STP.
Therefore, the volume of CO2 produced can be calculated as follows:

Volume of CO2 = (number of moles of CO2) * (22.4 dm^3 / mole)
= 1.4 moles * 22.4 dm^3/mole
= 31.36 dm^3 or 31.36 liters

So, the volume of carbon dioxide given off at STP when 20 grams of octane is burned in excess air is approximately 31.36 dm^3 or 31.36 liters.

Now, let's move on to deducing the formula of a hydrocarbon with a general formula of CnH2n-2 with a molar mass of 110 g/mol.

To deduce the formula, we need to follow these steps:

Step 1: Determine the molar mass of the hydrocarbon.
The molar mass of a hydrocarbon with a general formula of CnH2n-2 is given as 110 g/mol.

Step 2: Assign variables to the unknowns.
Let's assign "C" as the number of carbon atoms and "H" as the number of hydrogen atoms.

Step 3: Set up the equation based on the given formula.
The molar mass of the hydrocarbon can be calculated using the atomic masses of carbon and hydrogen.

Molar mass of the hydrocarbon = (C * atomic mass of carbon) + (H * atomic mass of hydrogen)

Step 4: Solve the equation.
Based on the equation, we have:
110 g/mol = (C * 12.01 g/mol) + (H * 1.01 g/mol)

Step 5: Simplify and solve for variables.
110 = 12.01C + 1.01H

To find the possible values of C and H, we can use trial and error by substituting different values for C and solving for H. We need to find values that satisfy the equation and result in whole number values for C and H.

For example, let's try C = 7:
110 = 12.01(7) + 1.01H
110 = 84.07 + 1.01H
25.93 = 1.01H
H ≈ 25.67 (not a whole number)

After trying different values of C, we find that when C = 6 and H = 12, the equation is satisfied:
110 = 12.01(6) + 1.01(12)
110 = 72.06 + 12.12
110 = 84.18 (approximation due to rounding)

Therefore, when C = 6 and H = 12, the formula of the hydrocarbon is C6H12.

In conclusion, the formula for the hydrocarbon with a general formula of CnH2n-2 and a molar mass of 110 g/mol is C6H12.