How do you balance $P_{4} + B r_{2} \to P B r_{3}$ $P_4 + Br_2 -> PBr_3$ ?

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How do you balance $P_{4} + B r_{2} \to P B r_{3}$ $P_4 + Br_2 -> PBr_3$ ?

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Answer 1 · 2016-03-07T01:58:11

We must make sure the number of molecules in each element on each side of the equation are equal.

Explanation:

The lowest number of $P_{4}$ $P_4$ that we can have is $1 P_{4}$ $1P_4$ , which means one molecule of 4 atoms of phosphorus.

This means we need 4 P atoms on the other side. We do this by making the equation:

$1 P_{4} + B r_{2} \to 4 P B r_{3}$ $1P_4 + Br_2 -> 4PBr_3$

This means that we have ( $4 \times 3 = 12)$ $4 xx 3 = 12)$ atoms of bromine. To finish the balancing process, we need to balance the bromine on the other side of the equation. It would be $6 b r_{2}$ $6br_2$ to make 12 atoms of bromine.

So, your equation will be $P_{4} + 6 B r_{2} \to 4 P B r_{3}$ $P_4 + 6Br_2 -> 4PBr_3$

Hopefully this helps!

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How do you balance $P_{4} + B r_{2} \to P B r_{3}$ $P_4 + Br_2 -> PBr_3$ ?

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How do you balance $P_{4} + B r_{2} \to P B r_{3}$ $P_4 + Br_2 -> PBr_3$ ?