If 30mL of H2 and 20 mL of O2 reacts to form water, what is left at the end of the reaction?

Explanation

Here's how to solve the problem:Write the balanced chemical equation:2 H₂ + O₂ → 2 H₂OConvert volumes to moles:Use the ideal gas law (PV = nRT) or assume ideal gas behavior at STP to convert volumes to moles:For H₂: moles = volume / molar volume = 30 mL / 22.4 L/mol ≈ 0.00134 molFor O₂: moles = volume / molar volume = 20 mL / 22.4 L/mol ≈ 0.00893 molDetermine the limiting reactant:Compare the mole ratios of H₂ and O₂ to their stoichiometric coefficients in the balanced equation (2:1).Based on the mole ratio, there is more O₂ (0.00893 mol) than required for the available H₂ (0.00134 mol). This means H₂ is the limiting reactant.Reactants consumed and products formed:Since H₂ is the limiting reactant, all 0.00134 mol of H₂ will react.According to the balanced equation, each mole of H₂ reacts with 0.5 mol of O₂ to form 1 mol of H₂O.Therefore, 0.00134 mol of H₂ will consume 0.5 * 0.00134 mol = 0.00067 mol of O₂.The remaining O₂ (0.00893 mol - 0.00067 mol) = 0.00826 mol will not react and will be left over.Convert moles of O₂ remaining to volume:Volume of remaining O₂ = moles * molar volume = 0.00826 mol * 22.4 L/mol ≈ 5 mLTherefore, 5 mL of O₂ will be left at the end of the reaction.