Question

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what volume of bromine is produced when 127 liters of bromine trifluoride react according to the following reaction? (all gases are at the same temperature and pressure.)
bromine trifluoride(g)→bromine(g)+fluorine(g)
liters bromine
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Explanation:

Step1: Write the balanced chemical equation

$$2BrF_3(g) ightarrow Br_2(g)+3F_2(g)$$

From the equation, the mole ratio of \(BrF_3\) to \(Br_2\) is \(2:1\).

Step2: Use Avogadro's law (at same \(T\) and \(P\), \(V\propto n\))

If \(V_1\) is the volume of \(BrF_3\) (\(V_1 = 127\) L) and \(V_2\) is the volume of \(Br_2\).
Since \(\frac{n_1}{n_2}=\frac{V_1}{V_2}\) and \(\frac{n_1}{n_2}=\frac{2}{1}\) (from mole - ratio), then \(\frac{V_1}{V_2}=\frac{2}{1}\).
So \(V_2=\frac{V_1}{2}\).

Step3: Calculate the volume of \(Br_2\)

Substitute \(V_1 = 127\) L into \(V_2=\frac{V_1}{2}\), we get \(V_2=\frac{127}{2}=63.5\) L.

Answer:

\(63.5\) liters bromine