IE: A sample of 7.75 g of Mg(OH)2 is added to 25.0 ml of 0.200 M HN03. (a) Write the chemical equation for the reaction that occurs, (b) Which is the limiting reactant in the reaction? (c) How many moles of Mg(OH)2, HN03, and Mg(N03)2 are present after the reaction is complete?
Solution 109 IE: Step1: a) Magnesium hydroxide reacts with nitric acid to form magnesium nitrate and water. The chemical equation for the reaction is: Mg(OH) (aq2+ 2HNO (aq) --3--> Mg(NO ) (aq) + 2H3 2) 2 Step2: b) Given, mass of Mg(OH) added = 725 g Molar mass of Mg(OH) = 58.322/mol Therefore, number of moles of Mg(OH) = mass of Mg(OH) / molar mass 2 7.75 2 = 58.32 = 0.133 mol Again, molarity of the HNO soluti3 = 0.200 M Volume of the HNO solution = 25.0 mL = 25 L = 0.025 L 3 1000 Therefore, number of moles of HNO solution =3olume of solution in litres X Molarity = 0.025 L X 0.200 M = 0.005 mol. Step3: From the equation above, it is clear that, 2 mol of HNO reacts with3 mol of Mg(OH) 2 for complete neutralization. Therefore, 0.005 mol of HNO reacts with (0.005 / 2 = 0.0025) mol of Mg(OH) for complete 3 2 neutralization. It is seen that all the nitric acid is consumed and magnesium hydroxide is present in excess. Thus, nitric acid is the limiting reagent in the above reaction. Step4: d) Since all the nitric acid is consumed, therefore, 0 mol of nitric acid is present after the reaction is complete. Number of moles of magnesium hydroxide remained = number of mol of magnesium hydroxide befor reaction - number of mol of magnesium hydroxide reacted. = 0.133 - 0.0025 = 0.1305 mol Number of moles of magnesium nitrate formed = number of moles of magnesium hydroxide reacted = 0.0025 mol.