Consider the combustion reaction of propane: C₃H₈(g)+ 5 O₂(g)$\rightarrow$3 CO₂(g)+ 4 H₂O(g), where $\Delta$H = -531 kcal. If 6.70 × 10⁴ kcal of energy is released in the reaction, how many grams of oxygen were consumed?

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat released in the reaction:
531 kcal/mol × 1 mol of propane = 531 kcal/mol
Next, we can use the equation q = mcΔT to calculate the amount of energy produced when 1 mol of propane is burned:
6.70 × 1014 kcal × (1 mol/propane) = 6.70 × 1014 kcal/mol
Now we can set up a proportion to determine the number of moles of oxygen consumed:
5 moles of O2 / 1 mole of propane = x moles of O2 / 6.70 × 1014 kcal
x = (6.70 × 1014 kcal)(5 mol O2 / 1 mol propane) = 3.35 × 1015 kcal
Finally, we can convert the number of moles of oxygen to grams using the molar mass of O2:
3.35 × 1015 mol O2 × (32 g/mol) = 1.07 × 1018 g of O2
However, the question is asking for the amount of oxygen consumed in grams to two significant figures, so we need to round to two significant figures and use scientific notation:
1.1 × 1018 g of O2 = 2.02 × 1014 g of O2. Therefore, the correct choice is (B).

Consider the Combustion Reaction of Propane: C3H8(g)+ 5 O2(g) →\rightarrow→ 3 CO2(g)+ 4 H2O(g), Where

Consider the Combustion Reaction of Propane: C₃H₈(g)+ 5 O₂(g) $\rightarrow$ 3 CO₂(g)+ 4 H₂O(g), Where