Write and balance the combustion equation for propane. (Propane is combusted in the presence of oxygen to produce carbon dioxide and water). 2) How many grams of oxygen are required to burn 200 grams of propane

363.64g of oxygen would be required.

Explanation:

1) Write and balance the combustion equation for propane.

Propane + Oxygen --> carbon dioxide + water

C3H8 + O2 --> CO2 + H2O

Upon balancing, we have;

C3H8 + 5O2 --> 3CO2 + 4H2O

2) How many grams of oxygen are required to burn 200 grams of propane.

From the reaction;

Propane = (3 * 12) + (8 * 1) = 44

Oxygen = (5 * 16) = 80

80 grams of oxygen is required to combust 44g of propane.

80 = 44

x = 200

x = ( 80 * 200 ) / 44

x = 363.64g

C3H8 + 5O2 → 3CO2 + 4H2O

We need 725.6 grams O2 to burn 200 grams of propane

Explanation:

Step 1: Data given

propane = C3H8

Molar mass propane = 44.1 g/mol

Mass of propane = 200.0 grams

Molar mass of O2 = 32.0 g/mol

Step 2: The balanced equation

C3H8 + 5O2 → 3CO2 + 4H2O

Step 3: Calculate moles propane

Moles propane = mass propane / molar mass propane

Moles propane = 200.0 grams / 44.1 g/mol

Moles propane = 4.535 moles

Step 4: Calculate moles O2

For 1 mol C3H8 we need 5 moles O2 to react to produce 3 moles CO2 and 4 moles H2O

For 4.535 moles propane we need 5*4.535 = 22.675  moles O2

Step 5: Calculate mass O2

Mass O2 = moles O2 * molar mass O2

Mass O2 = 22.675 moles * 32.0 g/mol

Mass O2 = 725.6 grams

We need 725.6 grams O2 to burn 200 grams of propane

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa

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