# How much heat will be released when 1.48g of chlorine reacts with excess phosphorus according to the equation 2p+5cl2 2pcl5∆H=-886kj?

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anor277 Share
Feb 9, 2018

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Slightly contrived circumstances here....

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SO we gots $\frac{1}{2} {P}_{4} \left(s\right) + 5 C {l}_{2} \left(g\right) \rightarrow 2 P C {l}_{5}$ $\Delta {H}_{\text{rxn as written}}^{\circ} = - 886 \cdot k J \cdot m o {l}^{-} 1$

The stated terms of the reaction are very poor. Phosphorus is likely to be oxidized to $P C {l}_{3}$ or a lesser oxidation state in the presence of limited quantities of chlorine gas.

We assume that the reaction proceeds as written and a stoichiometric quantity of phosphorus reacts....

$\text{Moles of dichlorine} = \frac{1.48 \cdot g}{70.90 \cdot g \cdot m o {l}^{-} 1} = 0.0209 \cdot m o l$...

And given the stoichiometry we gets...

$\frac{1}{2} \times 0.0209 \cdot m o l \times \left(- 886 \cdot k J \cdot m o {l}^{-} 1\right) = - 9.25 \cdot k J \ldots$

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