What volume of oxygen will be left unreacted when a mixture of 100cm^{3} of hydrogen and 200cm,sup>3,/sup> of oxygen are exploded in a eudiometer?

Answer Details

The balanced chemical equation for the reaction between hydrogen and oxygen is: 2H_{2} + O_{2} → 2H_{2}O This means that two molecules of hydrogen react with one molecule of oxygen to produce two molecules of water. Using the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature, we can calculate the number of moles of hydrogen and oxygen present in the mixture before the reaction. Assuming that the pressure and temperature are constant, we have: n_{H2} = (P * V_{H2}) / (R * T) = (1 atm * 100 cm^{3}) / (0.0821 L atm mol^{-1} K^{-1} * 298 K) = 0.00406 mol n_{O2} = (P * V_{O2}) / (R * T) = (1 atm * 200 cm^{3}) / (0.0821 L atm mol^{-1} K^{-1} * 298 K) = 0.00812 mol Since the stoichiometry of the reaction is 2:1 for hydrogen and oxygen, the limiting reactant is hydrogen. This means that all of the hydrogen will react with half of the oxygen to produce water, leaving the other half of the oxygen unreacted. The number of moles of water produced is given by: n_{H2O} = n_{H2} = 0.00406 mol The number of moles of oxygen consumed is given by: n_{O2} = n_{H2} / 2 = 0.00203 mol The number of moles of oxygen left unreacted is given by: n_{O2 unreacted} = n_{O2} - n_{O2 consumed} = 0.00812 mol - 0.00203 mol = 0.00609 mol Using the ideal gas law again, we can calculate the volume of oxygen left unreacted: V_{O2 unreacted} = (n_{O2 unreacted} * R * T) / P = (0.00609 mol * 0.0821 L atm mol^{-1} K^{-1} * 298 K) / 1 atm = 0.150 L = 150 cm^{3} Therefore, the volume of oxygen left unreacted is 150 cm^{3}.