220k views
1 vote
In a reaction involving the combustion of propane, the rate of consumption of propane was measured to be 0.6 mol/L s. What rate was carbon dioxide being formed at?

User Prabhakar
by
8.6k points

1 Answer

3 votes

Answer:

1.8 moles per liter of solution per second.

Step-by-step explanation:

The rate at which carbon dioxide is being formed in this reaction can be calculated by using the balanced chemical equation for the combustion of propane:


C3H8 + 5 O2 - > 3 CO2 + 4 H2O

From the balanced chemical equation, it is clear that for every mole of propane consumed, 3 moles of carbon dioxide are formed. Therefore, if the rate of consumption of propane is 0.6 mol/L s, the rate of formation of carbon dioxide is 3 * 0.6 = 1.8 mol/L s.

This means that the rate of formation of carbon dioxide is 1.8 moles per liter of solution per second.

User Corey Cole
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.