How do you balance $P_{4} (s) + C l_{2} (g) \to P C l_{5} (g)$ $P_4(s) + Cl_2(g) -> PCl_5(g)$ ?

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How do you balance $P_{4} (s) + C l_{2} (g) \to P C l_{5} (g)$ $P_4(s) + Cl_2(g) -> PCl_5(g)$ ?

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Answer 1 · 2017-06-12T02:02:17

$P_{4 (s)} + 10 C l_{2 (g)} \to 4 P C l_{5 (g)}$ $P_(4(s)) + 10Cl_(2(g)) -> 4PCl_(5(g))$

Explanation:

We'll balance the most complex compound first: $P C l_{5}$ $PCl_5$

The easiest way to do this is to balance the $P$ $P$ since the ratio of $P : P C l_{5}$ $P:PCl_5$ in $P C l_{5}$ $PCl_5$ is $1 : 1$ $1:1$ . On our reactants side, we have $P_{4}$ $P_4$ and thus we need a minimum of 4 $P$ $P$ on the products side to balance this. We'll make the coefficient of $P C l_{5}$ $PCl_5$ 4 to do this.

$P_{4 (s)} + C l_{2 (g)} \to 4 P C l_{5 (g)}$ $P_(4(s)) + Cl_(2(g)) -> 4PCl_(5(g))$

Now our phosphorus atoms are balanced. We see that in the reactants we have 2 $C l$ $Cl$ atoms ( $C l_{2}$ $Cl_2$ being diatomic), and in the products we have $4$ $4$ x $5 = 20 C l$ $5=20 Cl$ atoms (4 molecules of $P C l_{5}$ $PCl_5$ , 5 $C l$ $Cl$ atoms per molecule). By making the coefficient of $C l_{2}$ $Cl_2$ in the reactants side 10, we end up with 20 $C l$ $Cl$ atoms on each side. Overall, we now have 4 $P$ $P$ and 20 $C l$ $Cl$ on the reactants side, and 4 $P$ $P$ and 20 $C l$ $Cl$ on the products side. The equation has been balanced.

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How do you balance $P_{4} (s) + C l_{2} (g) \to P C l_{5} (g)$ $P_4(s) + Cl_2(g) -> PCl_5(g)$ ?

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Explanation:

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How do you balance P4(s)+Cl2(g)→PCl5(g)P_4(s) + Cl_2(g) -> PCl_5(g)?

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How do you balance $P_{4} (s) + C l_{2} (g) \to P C l_{5} (g)$ $P_4(s) + Cl_2(g) -> PCl_5(g)$ ?