Answer:
101.62 grams of water is given o when 62.1 g of propane burns.
Step-by-step explanation:
Propane gas (C₃H₈) burns with oxygen gas and products carbon dioxide gas and liquid water. The balanced reaction is:
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
By stoichiometry, the following amounts of moles of each compound participate in the reaction:
- C₃H₈: 1 mole
- O₂: 5 moles
- CO₂: 3 moles
- H₂O: 4 moles
Being the molar mass of each compound:
- C₃H₈: 44 g/mole
- O₂: 32 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
By stoichiometry, the following mass quantities of each compound participate in the reaction:
- C₃H₈: 1 mole* 44 g/mole= 44 grams
- O₂: 5 moles* 32 g/mole= 160 grams
- CO₂: 3 moles* 44 g/mole= 132 grams
- H₂O: 4 moles* 18 g/mole= 72 grams
Then you can apply the following rule of three: if by stoichiometry 44 grams of propane produces 72 grams of water, 62.1 grams of propane how much mass of water does it produce?
mass of water= 101.62 grams
101.62 grams of water is given o when 62.1 g of propane burns.