The combustion of propane may be described by the chemical equation c3h8(g)+5o2(g)⟶3co2(g)+4h2o(g) c3h8(g)+5o2(g)⟶3co2(g)+4h2o(g) how many grams of o2(g)o2(g) are needed to completely burn 83.3 g c3h8(g)?

To completely burn 83.3 grams of propane, there is 302.24 grams O2 needed.

Explanation:

Step 1: Data given

Mass of C3H8 = 83.3 grams

Molar mass of C3H8 = 44.1 g/mol

Molar mass of O2 = 32 g/mol

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g)

Step 3: Calculate moles of C3H8

Moles C3H8 = mass of C3H8 / Molar mass C3H8

Moles C3H8 = 83.3 grams / 44.1 g/mol

Moles C3H8 = 1.889 moles

Step 4: Calculate moles O2

For 1 mole C3H8 consumed, we need 5 moles of O2

For 1.889 moles of C3H8, we need 5* 1.889 = 9.445 moles of O2

Step 5: Calculate mass of O2

Mass O2 = moles O2 * Molar mass O2

Mass O2 = 9.445 moles * 32 g/mol

Mass O2 = 302.24 grams O2

To completely burn 83.3 grams of propane, there is 302.24 grams O2 needed.

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.