Gaseous butane CH3CH22CH3

Question

Gaseous butane CH3CH22CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O - Suppose 1.74 g of butane is mixed with 4.1 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Write the balanced equation for the reaction taking place:

2C₄H₁₀ + 13O₂ ==> 8CO₂ + 10H₂O ... balanced equation

Since we are given the amounts of BOTH reactants, we MUST find the limiting reactant. One way is to divide the moles of each reactant by the corresponding coefficient:

For C₄H₁₀: 1.74 g x 1 mol / 58.1 g = 0.0299 mols butane (÷2->0.015)

For O₂: 4.1 g O2 x 1 mol / 32 g = 0.128 mols oxygen (÷13->0.0098)

Since 0.0098 is less than 0.015, OXYGEN is limiting

Now, use the MOLES of O₂ to determine the maximum mass of CO₂.

0.128 mols O₂ x 8 mols CO₂ / 13 mols O₂ x 44 g CO₂ / mol = 3.5 g CO₂ (to 2 significant digits)