What is the limiting reactant of 4000 grams of heptane is combusted with 7000 grams of oxygen?
Well, I guess one might say that the limiting reactant is like the party pooper of the reaction. It's the one that gets used up first and stops the fun. So, in this case, let's analyze the situation.
To determine the limiting reactant, we need to compare the number of moles of each reactant. One mole of heptane weighs around 100 grams, and one mole of oxygen weighs approximately 32 grams.
So, for the 4000 grams of heptane, we would have approximately 40 moles. And for the 7000 grams of oxygen, we would have around 219 moles.
Now, when we check the balanced chemical equation, we see that the stoichiometric ratio between heptane and oxygen is 2:11.
Doing a little math, we find that if we have 40 moles of heptane, we would need 220 moles of oxygen to react completely. But since we only have 219 moles of oxygen, it means that oxygen is the limiting reactant, and heptane is left with its party hat on, waiting for more oxygen to arrive.
So, to answer your question, the limiting reactant in this case is oxygen, leaving heptane feeling a little breathless.
To determine the limiting reactant, we first need to balance the chemical equation for the combustion of heptane:
C7H16 + 11O2 -> 7CO2 + 8H2O
The balanced equation tells us that for every 1 mole of heptane (C7H16) reacted, 11 moles of oxygen (O2) are required.
Now, we need to convert the given masses of heptane and oxygen to moles. The molar mass of heptane (C7H16) is approximately 100.2 g/mol, and the molar mass of oxygen (O2) is approximately 32.0 g/mol.
Moles of heptane = 4000 g / 100.2 g/mol = 39.88 mol
Moles of oxygen = 7000 g / 32.0 g/mol = 218.75 mol
Next, we compare the mole ratio between heptane and oxygen. From the balanced equation, we know that 1 mole of heptane reacts with 11 moles of oxygen.
Mole ratio = moles of heptane / moles of oxygen
Mole ratio = 39.88 mol / 11 mol = 3.62
This means that for every 3.62 moles of heptane, 11 moles of oxygen are required.
Since the mole ratio is lower than the stoichiometric ratio of 1:11 (heptane:oxygen), the limiting reactant is heptane. Therefore, the limiting reactant is heptane in this scenario.
To determine the limiting reactant, we need to compare the number of moles of heptane and oxygen. The reactant that produces fewer moles of product is the limiting reactant.
1. Convert the masses of heptane and oxygen to moles.
- The molar mass of heptane (C7H16) is 100.2 g/mol.
- The molar mass of oxygen (O2) is 32 g/mol.
Moles of heptane = mass of heptane / molar mass of heptane
= 4000 g / 100.2 g/mol
Moles of oxygen = mass of oxygen / molar mass of oxygen
= 7000 g / 32 g/mol
2. Calculate the mole ratio between heptane and oxygen. The balanced chemical equation for the combustion of heptane is:
C7H16 + 11O2 → 7CO2 + 8H2O
From the equation, we see that 1 mole of heptane reacts with 11 moles of oxygen.
3. Determine the limiting reactant.
- Divide the moles of each reactant by the corresponding stoichiometric coefficient (from the balanced equation).
- The reactant with the smaller resulting value is the limiting reactant.
Moles of heptane needed = 1 * (moles of oxygen / 11)
Moles of oxygen needed = 11 * (moles of heptane)
Compare the moles of heptane needed with actual moles of heptane:
- If moles of heptane needed <= actual moles of heptane, then heptane is the limiting reactant.
- If moles of heptane needed > actual moles of heptane, then oxygen is the limiting reactant.
4. Calculate the moles of heptane needed and compare it with the actual moles of heptane to determine the limiting reactant.
Write the equation and balance it.
Convert g of both heptane and oxygen to mols. mols = grams/molar mass
One at a time, convert heptane to mols CO2 (as if you had all the oxygen needed); then convert oxygen to mols CO2 (as if you had all the heptane needed). Compare the two values. The correct value will be the SMALLER value of CO2 and the reagent producing that value will be the limiting reagent.