Feliza C.

asked • 06/08/21# 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

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?

## 1 Expert Answer

Hello, Feliza,

I'll start with part b, since we should identify the limiting reagent before deciding how much CO_{2} could be produced. The balanced equation says we need 5 moles of O_{2} for every 1 mole of C_{3}H_{8}. That means if we start with 8 moles C_{3}H_{8}, we would need 8x5=40 moles of O_{2}. 4 moles won't suffice, so O_{2} is the limiting reagent. In fact, 4 moles O_{2} will only require (4/5) of a mole of C_{3}H_{8}, leaving 71/5 moles of C_{3}H_{8}.

Now we can answer part a. The equation says we'll obtain 3 moles of CO_{2} for every 5 moles of O_{2}. That is a molar ratio of 3/5 moles CO_{2 }to moles O_{2}. Since we start with 4.0 moles of oxygen, we'll obtain (4.0 ~~moles O~~_{2})*(3 moles CO_{2 }/ 5 ~~moles O~~_{2}), or 12.0 moles CO_{2}, before all of the O_{2} is consumed, and the reaction stops.

We answered part c in part b. 7 1/5 moles of C_{3}H_{8} 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 H_{2}O produced. We'll then convert moles H_{2}O to grams.

Moles C_{3}H_{8}: 5.0 grams/44.0 g/mole = 0.114 moles C_{3}H_{8}

Moles O_{2}: 2.0 grams/32 g/mole = 0.0625 moles O_{2}

Remebering that we need 5 moles oxygen for every one mole C_{3}H_{8}, 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 C_{3}H_{8} consumed by using the molar ratio of C_{3}H_{8} to O_{2} of (1/5), amnd multiply it by the 0.0625 moles O_{2} that we now know will be completely consumed, taking(1/5)*(0.0625) or 0.0125 moles C_{3}H_{8} with it. This leaves 0.1011 moles of C_{3}H_{8} unreacted.

Next, let's find out how many moles are produced from the 0.0625 moles of oxygen. The molar ratio of H_{2}O to O_{2} 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 O_{2})*(4/5) = 0.05 moles H_{2}O.

The qurestion wants the amount of leftover reactant (C_{3}H_{8}) and water produced in grams, so multiply each mole answer by the molar mass of the respective compound:

(0.1011 moles C_{3}H_{8})*(44.03 g/mole) = 4.45 grams C_{3}H_{8}

(0.05 moles H_{2}O)*(18g/mole) = 0.90 grams H_{2}O

I hope this helps,

Bob

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Sujit S.

for questions b and c, you must specify the masses or reactants taken for reaction.06/08/21