91Ó°ÊÓ

Write the balanced chemical equation for the reaction between sulfur dioxide (SO2) and oxygen (O2) to form sulfur trioxide (SO3): \[2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)\]

The mole ratio between the reactants and products in the balanced chemical equation is: 2 moles of SO2 reacts with 1 mole of O2 to produce 2 moles of SO3.

The Ideal Gas Law is given by the equation: \(PV = nRT\) Since we are dealing with the same temperature (T) and pressure (P), we can use a simplified form of the Ideal Gas Law relating the initial and final volume (V) to the number of moles (n) and the gas constant (R). Thus, we have two equations: Initial Volume (V1) = (n_SO2 + n_O2)RT Final Volume (V2) = n_SO3RT

Divide V2 by V1 to find the ratio of final volume to initial volume: \(\frac{V2}{V1} = \frac{n_{SO3}RT}{(n_{SO2} + n_{O2})RT}\) Now we can cancel out the T and R terms from both numerator and denominator, and substitute the mole ratios from the balanced chemical equation (Step 2): \(\frac{V2}{V1} = \frac{2n_{SO2}}{2n_{SO2} + n_{O2}}\) Since equal moles of sulfur dioxide and oxygen have been mixed, we can say that: \(n_{SO2} = n_{O2} = n\) Substitute this into our volume ratio equation and simplify: \(\frac{V2}{V1} = \frac{2n}{2n + n} = \frac{2n}{3n}\) Now, cancel out the n term from both numerator and denominator: \(\frac{V2}{V1} = \frac{2}{3}\) Thus, the ratio of the final volume of the gas mixture to the initial volume of the gas mixture is 2:3.