Question

How many grams of oxygen are required to burn 13.5 g of acetylene

Answer 1

Answer:

41.54 grams of oxygen are required to burn 13.5 g of acetylene

Explanation:

The balanced reaction is:

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

• C₂H₂: 2 moles
• O₂: 5 moles
• CO₂: 4 moles
• H₂O: 2 moles

Being the molar mass of the compounds:

• C₂H₂: 26 g/mole
• O₂: 32 g/mole
• CO₂: 44 g/mole
• H₂O: 18 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

• C₂H₂: 2 moles* 26 g/mole= 52 grams
• O₂: 5 moles* 32 g/mole= 160 grams
• CO₂: 4 moles* 44 g/mole= 176 grams
• H₂O: 2 moles* 18 g/mole= 36 grams

You can apply the following rule of three: if by stoichiometry 52 grams of acetylene react with 160 grams of oxygen, 13.5 grams of acetylene react with how much mass of oxygen?

$mass of oxygen=\frac{13.5 grams of acetylene*160 grams of oxygen}{52 grams of acetylene}$

mass of oxygen= 41.54 grams

41.54 grams of oxygen are required to burn 13.5 g of acetylene