3. How many grams of oxygen are needed for the complete combustion of 11 grams of propane, C3H8?
C3H8 + 5 O2 + 3 CO2 + 4 H2O

Mass = 40 g

Explanation:

Given data:

Mass of oxygen needed = ?

Mass of propane = 11 g

Solution:

Chemical equation:

C₃H₈ +5O₂   →     3CO₂ + 4H₂O

Number of moles of propane:

Number of moles = mass/molar mass

Number of moles = 11 g/ 44.1 g/mol

Number of moles = 0.25 mol

now we will compare the moles of propane and oxygen.

              C₃H₈        :          O₂

                  1           :            5

                0.25      :            5/1×0.25 = 1.25 mol

Mass of oxygen needed:

Mass = number of moles × molar mass

Mass = 1.25 mol × 32 g/mol

Mass = 40 g

*tilts head* sounds more like a statement

a electronegativity decreases beacuse the size of nucleus is bigger.