The equation below shows the incomplete combustion of ethene.
C2H4(g) + 2O2(g) ==> 2CO(g) + 2H2O(g)
If 2.70 mol C2H4 is reacted with 6.30 mol O2,
a) identify the limiting reagent.
b) calculate the moles of water produced.
Show all steps.
limiting reagent is c2h4
moles of water produced is 5.4
a) Well, let's determine the moles of CO and H2O that will be produced if all the C2H4 and O2 react. For every 2 moles of C2H4, we get 2 moles of CO and 2 moles of H2O. So, if we have 2.70 mol of C2H4, that means we'll have:
2.70 mol C2H4 x (2 mol CO / 2 mol C2H4) = 2.70 mol CO
And,
2.70 mol C2H4 x (2 mol H2O / 2 mol C2H4) = 2.70 mol H2O
Now, let's consider the O2. For every 2 moles of O2, we get 2 moles of CO and 2 moles of H2O. So, if we have 6.30 mol of O2, we'll have:
6.30 mol O2 x (2 mol CO / 2 mol O2) = 6.30 mol CO
And,
6.30 mol O2 x (2 mol H2O / 2 mol O2) = 6.30 mol H2O
Since the stoichiometry says we should have a 1:1 ratio of CO and H2O for both reactants, the reactant that produces less product will be the limiting reagent. In this case, the limiting reagent is C2H4 since it produces fewer moles of CO and H2O.
b) Now that we know C2H4 is the limiting reagent, we can determine the moles of H2O produced. From our calculations in part a, we found that 2.70 mol of C2H4 produces 2.70 mol of H2O. So, the moles of H2O produced in this reaction will also be 2.70 mol. That's a lot of water! It's raining moles!
To identify the limiting reagent and calculate the moles of water produced, we need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced chemical equation.
a) To identify the limiting reagent:
Step 1: Write down the balanced chemical equation:
C2H4(g) + 2O2(g) ==> 2CO(g) + 2H2O(g)
Step 2: Convert the given number of moles of each reactant to moles of another reactant using the stoichiometric ratio from the balanced equation:
2.70 mol C2H4 x (2 mol O2 / 1 mol C2H4) = 5.40 mol O2
6.30 mol O2 stays the same
Step 3: Compare the moles of each reactant to determine which one is present in excess. The reactant that produces fewer moles of product is the limiting reagent.
The ratio of moles of C2H4 to moles of O2 is 2.70 to 5.40, which simplifies to 1 to 2.
The ratio of moles of O2 to moles of C2H4 is 6.30 to 6.30, which simplifies to 1 to 1.
Comparing the ratios, we can see that the O2 is present in excess because it is in a 2 to 1 ratio with the C2H4, while the C2H4 is in a 1 to 1 ratio with the O2. Therefore, the limiting reagent is C2H4.
b) To calculate the moles of water produced:
Step 1: Determine the moles of the limiting reagent used based on the given number of moles:
2.70 mol C2H4 x (2 mol H2O / 1 mol C2H4) = 5.40 mol H2O
Step 2: Since the stoichiometric ratio is 2 moles of H2O per 1 mole of C2H4, the number of moles of water produced is equal to the number of moles of the limiting reagent used, which is 5.40 mol.
Therefore, the limiting reagent is C2H4, and the number of moles of water produced is 5.40 mol.
Limiting reagent problems are pretty much alike. Print these instructions and memorize them.
1. Write and balance the equation. You have that.
2. Convert grams to moles. moles = grams/molar mass. Your problem has already done that but its a step most problems require.
3a. Using the coefficients in the balanced equation, convert moles ethene to moles of the product.
3b. Same procedure, convert moles oxygen to moles product.
3c. It is quite likely that the answer to 3a and 3b will not be the same. Obviously, one of them is wrong; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
4. Now use the value from 3c to convert to grams of the product. g = moles x molar mass.