which of the following would be the limited reagent in the reaction shown below there are 50 molecules of O2 and 50 molecules of H2
You need to do the work for this type problem and learn to do them.
1. Write the balanced equation.
2H2 + O2 ==> 2H2O
2. Convert what you have to mols. Molecules will do so go with what you have.
3. Using the coefficients in the balanced, convert molecules O2 and molecules H2 to molecules H2O. Two separate conversions; i.e., one for H2 and one for O2.
4. The one that produces the fewer molecules of H2O is the limiting reagent.
hahaha i wish i new how to do all that stuff im not very smart but ill try it ill prolly get it wrong anyway but thank you very much
Do what you can and repost what you are confused about. Be sure and explain what it is you don't understand. We can help you through it.
To determine the limiting reagent, we need to compare the amount of each reactant to the stoichiometry of the reaction.
In the reaction provided, there are 50 molecules of O2 and 50 molecules of H2.
The balanced equation for the reaction is:
2H2 + O2 -> 2H2O
According to the balanced equation, 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O.
To determine the limiting reagent, we need to convert the amounts of reactants to moles.
Number of moles of O2 = Number of molecules / Avogadro's number
= 50 / 6.022 × 10^23 molecules/mol
≈ 8.31 × 10^-23 moles
Number of moles of H2 = Number of molecules / Avogadro's number
= 50 / 6.022 × 10^23 molecules/mol
≈ 8.31 × 10^-23 moles
Now, let's compare the moles of each reactant to the stoichiometry of the reaction:
For O2: 8.31 × 10^-23 moles
For H2: 8.31 × 10^-23 moles
According to the stoichiometry, every mole of O2 reacts with 2 moles of H2.
Since the number of moles of H2 is equal to the number of moles of O2, there is an equal number of moles of each reactant.
Therefore, neither O2 nor H2 is present in a limiting amount. Both reactants are in excess, meaning the reaction will proceed completely until one of the reactants is completely consumed.