Ask AI: Na2H2EDTA.2H2O + FeCl3.6H2O + NaOH -> NA[Fe(EDTA)].3H20 + NaCl + H20 balanced equation and limiting reagents

First, let's balance the equation. Na2H2EDTA.2H2O + 2FeCl3.6H2O + 5NaOH -> 2Na[Fe(EDTA)].3H2O + 6NaCl + 8H2O This equation shows that one mole of Na2H2EDTA can react with two moles of FeCl3 to produce two moles of Na[Fe(EDTA)]. To find out the limiting reagent, you'll need to know how many moles of each reagent you have. You could determine the number of moles by dividing the mass of each reagent by its molecular weight. For example, if you have 100 g of Na2H2EDTA, 200 g of FeCl3 and 80 g of NaOH, you could calculate the number of moles as follows: Number of moles of Na2H2EDTA = 100 g / 372.24 g/mol = 0.27 mol Number of moles of FeCl3 = 200 g / 270.30 g/mol = 0.74 mol Number of moles of NaOH = 80 g / 39.99 g/mol = 2.00 mol Now compare the molar ratios of your reagents with the stoichiometric ratios in the balanced equation: Na2H2EDTA : FeCl3 : NaOH = 0.27 mol : 0.74 mol : 2.00 mol From the balanced equation, Na2H2EDTA : FeCl3 : NaOH = 1 mol : 2 mol : 5 mol From comparing these ratios, NaOH is your limiting reagent because in proportion to the other reagents, it is not enough to fully react with both Na2H2EDTA and FeCl3 considering the stoichiometric molar ratio in the balanced chemical equation. Therefore, NaOH will be used up first and the reaction will stop, making it the limiting reagent. Please note that this example assumes the amounts given. If you have different amounts of the substances, you need to recalculate the moles and compare the ratios again.