Using the equation, C5H12 + 8O2 Imported Asset 5CO2 + 6H2O, if an excess of pentane (C5H12) were supplied, but only 4 moles of oxygen were available, how many moles of water would be produced?

the balanced equation for the combustion of pentane is as follows:

C5H12 + 8O2 - - > 5CO2 + 6H2O

excess pentane is allowed to react with 4 moles of oxygen. this means that pentane is the excess reactant and O ₂ is the limiting reactant. Limiting reactant is the reagent that is fully used up in the reaction and amount of product formed depends on the amount of limiting reactant present.

Stoichiometry of O ₂ to H ₂ O is 8: 6.

If 8 mol of O ₂ forms 6 mol of H ₂ O then 4 mol of O ₂ forms - 6/8 x 4 = 3 mol of H ₂ O

therefore 3 mol of H ₂ O is formed

Answer: 3 moles of water would be produced in present case.

Reason:

Reaction involved in present case is:

C5H12 + 8O2 → 5CO2 + 6H2O

In above reaction, 1 mole of C5H12 reacts with 8 moles of oxygen to give 6 moles of water.

Thus, 4 moles of oxygen will react with 0.5 mole of C5H12, to generate 3 moles of H2O.