The balanced equation is C3H8+ 5O2 --------> 3CO2+4H2O, if 8.0 mol C3H8 reacts with 4.0 mol O2. Please show the work for A-E

Question

a. how many moles of CO2 will be produced?

b. what is the limiting reactant? the excess reactant?

c. how many moles of excess reactant were left unreacted?

d. if 5.0 grams of C3H8 reacts with 2.0 grams of O2, how many grams of H2O will be produced?

e. how many grams are unreacted?

Answer 1

Hello, Feliza,

I'll start with part b, since we should identify the limiting reagent before deciding how much CO₂ could be produced. The balanced equation says we need 5 moles of O₂ for every 1 mole of C₃H₈. That means if we start with 8 moles C₃H₈, we would need 8x5=40 moles of O₂. 4 moles won't suffice, so O₂ is the limiting reagent. In fact, 4 moles O₂ will only require (4/5) of a mole of C₃H₈, leaving 71/5 moles of C₃H₈.

Now we can answer part a. The equation says we'll obtain 3 moles of CO₂ for every 5 moles of O₂. That is a molar ratio of 3/5 moles CO₂ to moles O₂. Since we start with 4.0 moles of oxygen, we'll obtain (4.0 moles O₂)*(3 moles CO₂ / 5 moles O₂), or 12.0 moles CO₂, before all of the O₂ is consumed, and the reaction stops.

We answered part c in part b. 7 1/5 moles of C₃H₈ will remain. Zounds!

Part d requires conversion of the reactant masses into moles before we can take the same steps as above, except we will now look for moles H₂O produced. We'll then convert moles H₂O to grams.

Moles C₃H₈: 5.0 grams/44.0 g/mole = 0.114 moles C₃H₈

Moles O₂: 2.0 grams/32 g/mole = 0.0625 moles O₂

Remebering that we need 5 moles oxygen for every one mole C₃H₈, it is easy to see that we are limited by the amount of oxygen, again. The reaction will stop once the 0.0625 moles of oxygen is consumed. We can calculate the moles C₃H₈ consumed by using the molar ratio of C₃H₈ to O₂ of (1/5), amnd multiply it by the 0.0625 moles O₂ that we now know will be completely consumed, taking(1/5)*(0.0625) or 0.0125 moles C₃H₈ with it. This leaves 0.1011 moles of C₃H₈ unreacted.

Next, let's find out how many moles are produced from the 0.0625 moles of oxygen. The molar ratio of H₂O to O₂ is 4 to 5, or 4/5, according to the balanced equation. We can multiply this ratio by themoles oxygen consumed to find moles water produced: (0.0625 moles O₂)*(4/5) = 0.05 moles H₂O.

The qurestion wants the amount of leftover reactant (C₃H₈) and water produced in grams, so multiply each mole answer by the molar mass of the respective compound:

(0.1011 moles C₃H₈)*(44.03 g/mole) = 4.45 grams C₃H₈

(0.05 moles H₂O)*(18g/mole) = 0.90 grams H₂O

I hope this helps,

Bob