For the given reaction, what volume of O2 would be required to react with 5.8 L of SO2 , measured at the same temperature and pressure? 2SO2(g)+O2(g)⟶2SO3(g)

Answer 1

The ideal gas law (PV=nRT) relates number of moles of a gas, volume, pressure, and temperature. Fortunately in this problem the pressure and volume do not change, meaning that the mole ratio in the equation can be replaced with the volume ratio.

5.8 L SO2 x 1 L O2 / 2 L SO2 = 2.9 L O2

Notice the units are correct (L SO2 cancel out) and the number of sig figs is correct (2 in the given number, 2 in the answer).