Answer :
Answer:
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
Answer:
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