To determine the limiting reactant, we first need to find the number of moles of each reactant.The molar mass of ammonia NH3 is approximately 14 N + 3 * 1 H = 17 g/mol.The molar mass of oxygen O2 is approximately 2 * 16 O = 32 g/mol.Now, we can find the number of moles of each reactant:moles of NH3 = 100 g / 17 g/mol = 5.88 molmoles of O2 = 50 g / 32 g/mol = 1.56 molNext, we need to compare the mole ratios of the reactants to the balanced equation:moles of NH3 / 4 = 5.88 / 4 = 1.47moles of O2 / 3 = 1.56 / 3 = 0.52Since 0.52 is smaller than 1.47, oxygen O2 is the limiting reactant.Now, we need to find out how much ammonia NH3 will be left over after the reaction is complete. To do this, we can use the mole ratio from the balanced equation:moles of NH3 used = 4 moles NH3 / 3 moles O2 * moles of O2moles of NH3 used = 4/3 * 1.56 = 2.08 molNow, we can find the moles of NH3 left over:moles of NH3 left over = initial moles of NH3 - moles of NH3 usedmoles of NH3 left over = 5.88 - 2.08 = 3.80 molFinally, we can convert the moles of NH3 left over back to grams:grams of NH3 left over = moles of NH3 left over * molar mass of NH3grams of NH3 left over = 3.80 mol * 17 g/mol = 64.6 gSo, the limiting reactant is oxygen O2 , and there will be 64.6 grams of ammonia NH3 left over after the reaction is complete.