🤓 Based on our data, we think this question is relevant for Professor Li's class at UW-SEATTLE.
Liquid octane (CH3(CH2)6CH3) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). Suppose 2.28 g of octane is mixed with 4.5 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.
We’re being asked to calculate the mass of carbon dioxide, CO2, that will be formed in the reaction. First, we need to find the chemical equation:
CH3(CH2)6CH3(l) + O2(g) → CO2(g) + H2O(g)
This equation is not yet balanced. To balance it, we have to make sure that the number of elements on both sides is equal.
Balance C: We have 8 C on the reactant side and 1 C on the product side – add a coefficient of 8 to CO2:
CH3(CH2)6CH3(l) + O2(g) → 8 CO2(g) + H2O(g)
Balance H: We have 18 H on the reactant side and 2 H on the product side – add a coefficient of 9 to H2O:
CH3(CH2)6CH3(l) + O2(g) → 8 CO2(g) + 9 H2O(g)
Balance O: We have 2 O on the reactant side and 8(2) + 9 = 25 O on the product side – multiply the entire equation by 2 and add a coefficient of 25 to O2:
2 CH3(CH2)6CH3(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
Now that we have a balanced equation, we can determine how much CO2 is produced.
Notice that we are given the mass of both reactants: this means we need to determine the limiting reactant, which is the reactant that forms the less amount of product. This is because once the limiting reactant is all used up, the reaction can no longer proceed and make more products.
This means the limiting reactant determines the maximum mass of the product formed.