Can you please help me with chemistry?

Question

Acetylene C2H2  gas and oxygen O2 gas react to form carbon dioxide CO2 gas and water H2O

vapor. Suppose you have 13.0 mol of C2H2  and 11.0 mol of O2 in a reactor. Calculate the largest amount of 

CO2 that could be produced. Round your answer to the nearest 0.1 mol.

.

Answer 1

First, write the correctly balanced equation for the reaction:

2C₂H₂ + 5O₂ ==> 4CO₂ + 2H₂O ... balanced equation for combustion of acetylene

Since we are given moles of BOTH reactants, we must first find which reactant is limiting.

Simply divide moles of each by their respective coefficient to see which is limiting:

--- 13.0 mol C₂H₂ / 2 = 6.5

--- 11.0 mol O₂ / 5 = 2.2

Since 2.2 is < 6.5, this tells us that 11.0 mol O₂ is limiting and will run out first. It is the limiting reactant that determines how much product will form.

Largest amount of CO₂ that can form = 11.0 mol O₂ x 4 mol CO₂ / 5 mol O₂ = 8.8 moles CO₂