Given the equation below, how does the rate of disappearance of O2 compare to the rate of disappearance of C3H8?C3H8 + 5 O2 → 3 CO2 + 4 H2OQuestion 2Select one:a.O2 gets used up five times faster than C3H8.b.O2 gets used up one third as fast as C3H8.c.O2 gets used up three times faster than C3H8.d.O2 gets used up one fifth as fast as C3H8.
Question
Given the equation below, how does the rate of disappearance of O2 compare to the rate of disappearance of C3H8?C3H8 + 5 O2 → 3 CO2 + 4 H2OQuestion 2Select one:a.O2 gets used up five times faster than C3H8.b.O2 gets used up one third as fast as C3H8.c.O2 gets used up three times faster than C3H8.d.O2 gets used up one fifth as fast as C3H8.
Solution
The answer is a. O2 gets used up five times faster than C3H8.
Here's why:
The balanced chemical equation is C3H8 + 5 O2 → 3 CO2 + 4 H2O.
The coefficients in front of the chemical formulas represent the stoichiometric ratios of the reactants and products. In this case, for every 1 molecule of C3H8 that reacts, 5 molecules of O2 are required. Therefore, O2 is used up five times faster than C3H8.
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