The reaction of 2.5 mol of O2 with 4.6 mol of C3H8 will produce ________ mol of H2O.
This is a limiting reagent (LR) problem and we know that because amounts are given for BOTH reactants.
C3H8 + 5O2 ==> 3CO2 + 4H2O
mols H2O formed if we used 2.5 mol O2 and all of the propane we needed is 2.5 mol O2 x (4 mol H2O/5 mol O2) = 2.5 x 4/5 = 2 mol H2O formed.
mols H2O formed if we used 4.6 mol C3H8 and all of the O2 we needed That's 4.6 mol C3H8 x (4 mols H2O/1 mol C3H8) = 4.6 x 4/1 = 18.4
We have two answers for mols H2O formed and both of them can't be right. The correct answer in LR problems is ALWAYS the smaller value and the reagent providing that number is the LR. So 2 mols H2O will be formed.
Well, according to my calculations, the reaction between 2.5 mol of O2 and 4.6 mol of C3H8 will produce a molecular circus of 999.99 mol of H2O! That's a whole lot of water for a chemistry show! Just make sure to have some towels ready in case things get a little splashy.
To determine the number of moles of H2O produced, we need to use the balanced chemical equation for the reaction between O2 and C3H8. The balanced equation is:
C3H8 + 5O2 -> 3CO2 + 4H2O
From this equation, we can see that for every 1 mole of C3H8 reacted, 4 moles of H2O are produced. Therefore, we need to calculate the moles of C3H8 reacted, and then multiply that by the mole ratio of H2O to C3H8.
Given:
Moles of O2 = 2.5 mol
Moles of C3H8 = 4.6 mol
Since the mole ratio of C3H8 to O2 is 1:5, we can calculate the moles of O2 required for the reaction:
Moles of O2 = 5 * Moles of C3H8
2.5 mol = 5 * Moles of C3H8
Moles of C3H8 = 2.5 mol / 5
Moles of C3H8 = 0.5 mol
Now we can calculate the moles of H2O produced:
Moles of H2O = 4 * Moles of C3H8
= 4 * 0.5 mol
= 2 mol
Therefore, the reaction of 2.5 mol of O2 with 4.6 mol of C3H8 will produce 2 mol of H2O.
To determine the number of moles of H2O produced in the reaction between 2.5 mol of O2 and 4.6 mol of C3H8, we need to first determine the balanced chemical equation for the reaction.
The balanced equation for the combustion of C3H8 (propane) with O2 is:
C3H8 + 5O2 -> 3CO2 + 4H2O
From the balanced equation, we can see that for every 1 mole of C3H8, 4 moles of H2O are produced.
Therefore, to find the number of moles of H2O produced, we need to calculate the number of moles of C3H8 and then multiply it by the molar ratio of H2O to C3H8.
Given that there are 4.6 moles of C3H8, we can calculate the number of moles of H2O produced by multiplying it by the molar ratio:
4.6 mol C3H8 * (4 mol H2O / 1 mol C3H8) = 18.4 mol H2O
Therefore, the reaction of 2.5 mol of O2 with 4.6 mol of C3H8 will produce 18.4 mol of H2O.