Question

The combustion of 15.3g Acetylene, C₂H₂ in 39.4g of oxygen yields how much water? Label the limiting and excess reactants.

Please help! I don't get it.

Answer 1

Balanced chemical equation for the combustion of acetylene is given as:

2C2H2 + 5O2 4CO2 + 2H2O

Given mass of acetylene = 15.3 g

Molar mass of acetylene = 26 g/mol

So, moles of acetylene = mass / molar mass

= 15.3 g / 26 g/mol

= 0.665 moles

Given mass of oxygen = 39.4 g

Molar mass of oxygen (O2) = 32 g/mol

So, moles of oxygen = mass / molar mass

= 1.138 moles

From the Balanced chemical equation, it is clear that

2 moles of acetylene requires O2 for the combustion = 5 moles

Then, 0.665 moles of acetylene will require O2 for combustion = 5/2 × 0.665

= 1.663 moles

So, amount of oxygen required is 1.663 moles but available oxygen is 1.138 moles only. This means oxygen will be consumed completely in the reaction. ​​​​​​So, oxygen is the limiting reactant and acetylene is the excess reactant.

To calculate the amount of water produced:

From the Balanced chemical equation,

5 moles of O2 produces H2O = 2 moles

Then, 1.138 moles of O2 produces = 2/5 × 1.138

= 0.455 moles

So, moles of H2O produced is 0.455 moles

Molar mass of water = 18 g/mol

So, mass of H2O = molar mass × no. of moles

= 18 g/mol × 0.455 moles

= 8.19 g

Amount of H2O produced is 8.19 g