Liquid octane (CH3(CH2)CH3) reacts with gaseous oxygen gas(O2) to produce gaseous carbon dioxide(CO2) and gaseous water (H2O). What is the theoretical yield of carbon dioxide formed from the reaction of 27.4g of octane and 77.8g of oxygen gas?

The theoretical yield of carbon dioxide formed is 68.4 grams CO2

Explanation:

Step 1: Data given

Liquid octane = C8H18

gaseous oxygen gas = O2

gaseous carbon dioxide = CO2

gaseous water = H2O

Mass of octane = 27.4 grams

Molar mass of octane = 114.23 g/mol

Mass of oxygen = 77.8 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The reaction

2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g)

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles of octane = 27.4 grams / 114.23 g/mol

Moles of octane = 0.240 moles

Moles of oxygen = 77.8 grams / 32.0 g/mol

Moles of oxygen = 2.43 moles

Step 4: Calculate the moles of the products

For 2 moles of octane we need 25 moles of O2 to produce 16 moles of CO2 and 18 moles H2O

O2 is the limiting reactant. It will completely react. (2.43 moles).

There will react 2.43/12.5 = 0.194 moles

There will remain 0.240 - 0.194 =  0.046 moles of octane.

There will be produced:

16/25 * 2.43 = 1.555 moles of CO2

18/25 * 2.43 = 1.750 moles of H2O

This is:

1.56 moles * 44.01 g/mol = 68.4 grams CO2

1.750 moles * 18.02 g/mol = 31.5 grams H2O

The theoretical yield of carbon dioxide formed is 68.4 grams CO2

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.