Question

How much heat is produced when $2.63gm$ of phosphorous and $40gm$ of bromine is allowed to react according to the following equation?

$P_{4\left(s\right)}+6B{r}_{2\left(l\right)}\to 4PB{r}_{3\left(g\right)};\Delta H=-486kJ$

• A. -10.30 KJ
• B. -20.6 KJ
• C. -40.12 KJ
• D. -5.63 KJ

Answer 1

The correct option is A -10.30 KJ

Moles of $P_4\left(s\right)=\frac{2.63}{124}\simeq 0.02$
Moles of $B{r}_2=\frac{40}{80\times 2}$ $=\frac{1}{4}=0.25$
$therefore,P_4$ is a limiting reagent.
$\therefore Heat=-$486 x $\frac{2.63}{124}$ KJ $=-$ 10.3 KJ

Option A is correct.