3.2 grams of hydrogen react with 9.5 grams of bromine. Which is the limiting reagent?

Question

Noah Aldrich · Accepted Answer

Bromine.

The first thing to do here is write a balanced chemical equation that describes this synthesis reaction

#"H"_ (2(g)) + "Br"_ (2(g)) -> 2"HBr"_ ((g))#

Hydrogen gas will react with bromine gas at temperatures that exceed #300^@"C"# to form hydrogen bromide, #"HBr"#. Notice that the two reactants react in a #1:1# mole ratio.

This tells you that the reaction will always consume equal numbers of moles of hydrogen gas and bromine gas.

Use the molar masses of the two reactants to see how many moles of each you're mixing

#3.2 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.016color(red)(cancel(color(black)("g")))) = "1.5873 moles H"_2#
#9.5 color(red)(cancel(color(black)("g"))) * "1 mole Br"_2/(159.81color(red)(cancel(color(black)("g")))) = "0.05945 moles Br"_2#

As you can see, you have significantly fewer moles of bromine gas than needed in order to ensure that all the moles of hydrogen gas reacts.

That many moles of hydrogen gas would have required

#1.5873 color(red)(cancel(color(black)("moles H"_2))) * "1 mole Br"_2/(1color(red)(cancel(color(black)("mole H"_2)))) = "1.5873 moles Br"_2#

Since you don't have enough moles of bromine available, you can say for a fact that bromine gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of hydrogen gas will get a chance to react.

This is of course equivalent to saying that hydrogen gas is in excess.

Henry Antrim · Answer

undefined To determine the limiting reagent, we need to calculate the moles of each reactant. Then, we can use the stoichiometry of the balanced chemical equation to see which reactant runs out first.

First, we convert the masses of hydrogen and bromine to moles using their respective molar masses:
- Hydrogen (H2): 3.2 grams / (2 grams/mole) = 1.6 moles
- Bromine (Br2): 9.5 grams / (159.808 grams/mole) = 0.0596 moles

Next, we look at the balanced chemical equation for the reaction between hydrogen and bromine:
H2 + Br2 -> 2HBr

From the equation, we see that 1 mole of hydrogen reacts with 1 mole of bromine to produce 2 moles of hydrogen bromide.

Since 1 mole of bromine is needed for every 1 mole of hydrogen, and we have more moles of hydrogen than bromine, the limiting reagent is bromine.