# How do you identify the limiting reactant when 5.78 g of phosphorus react with 27.9 of liquid bromine?

## The reaction between phosphorus and liquid bromine is outlined as: $2 P \left(s\right) + 3 B {r}_{2} \left(l\right) \to 2 P B {r}_{3} \left(l\right)$

Nov 28, 2016

By taking the molar quantities of each reagent........

#### Explanation:

You have written the stoichiometric equation; we could use half-inegral coefficients to make the arithmetic a bit easier:

$P \left(s\right) + \frac{3}{2} B {r}_{2} \left(l\right) \rightarrow P B {r}_{3} \left(l\right)$

$\text{Moles of phosphorus}$ $=$ $\frac{5.78}{31.00 \cdot g \cdot m o {l}^{-} 1} = 0.181 \cdot m o l$,

$\text{Moles of bromine}$ $=$ $\frac{27.9}{159.8 \cdot g \cdot m o {l}^{-} 1} = 0.175 \cdot m o l$,

Clearly, the bromine is in stoichiometric deficiency. Why? Because stoichiometric equivalence would demand,

$0.181 \cdot m o l \times \frac{3}{2} \times 159.8 \cdot g \cdot m o l \cong 44 \cdot g$ bromine.

There are few things in the laboratory that are as nasty and corrosive as liquid bromine. It would cause horrendous burns, and it is treated with considerable respect.