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

Jul 30, 2016

Bromine.

#### Explanation:

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

${\text{H"_ (2(g)) + "Br"_ (2(g)) -> 2"HBr}}_{\left(g\right)}$

Hydrogen gas will react with bromine gas at temperatures that exceed ${300}^{\circ} \text{C}$ to form hydrogen bromide, $\text{HBr}$. 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.