Question

For the following reaction, 0.551 moles of butane (C4H10) are mixed with 0.119 moles of oxygen gas. butane (C4H10)(g) + oxygen(g) → carbon dioxide(g) + water(8) What is the formula for the limiting reagent? What is the maximum amount of carbon dioxide that can be produced? moles

Answer 1

The balanced equation is

2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O

From the balanced equation we can say that

2 mole of butane requires 13 mole of O2 so

0.551 mole of butane will require

= 0.551 mole of butane *(13 mole of O2 / 2 mole of butane)

= 3.58 mole of O2

But we have 0.119 mole of O2 which is in short so O2 is limiting reactant

Therefore, the limiting reactant is O₂

From the balanced equation we can say that

13 mole of O2 produces 8 mole of CO2 so

0.119 mole of O2 will produce

= 0.119 mole of O2 *(8 mole of CO2 / 13 mole of O2)

= 0.0732 mole of CO2

Therefore, the number of moles of CO2 produced would be 0.0732 mole