Question

How many grams of water, H 2 O , is given o when 62.1 g of propane, C 3 H 8 , burns?

Answer 1

Answer:

101.62 grams of water is given o when 62.1 g of propane burns.

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?

$mass of water=\frac{62.1 grams of propane*72 grams of water}{44 grams of propane}$

mass of water= 101.62 grams

101.62 grams of water is given o when 62.1 g of propane burns.