Question

3. In the reaction where 40.0g of sodium hydroxide reacts with 60.0g of sulfuric acid... Predict the products. Then, write the balanced chemical equation: Identify the type of reaction Determine the limiting reactant: Calculate the amount (g) of sodium sulfate formed: Calculate the amount (g) of reactant in excess that remains:

Answer 1

Ans. # Type of reaction: Acid-Base reaction

Given: Balanced reaction: In balanced reaction, theoretical molar ratio of reactant: Mass of H2S04 H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(l) Formation of product follows the stoichiometry of the limiting reactant. Step 2: Mol of H2SO4 1.0 Mol of NaOH 2.0 MW of H2SO4 = 9807984 g mol-' MW of NaOH 39.997108 g mol Following stoichiometry of balanced reaction, 2 mol NaOH forms 1 mol So, moles of Na2SO4 formed - 0.50004 mol Now 60.00 g 40.00 g Na2SO4 Mass of NaOH Step 1: Detemine lmiting reactant, if any Mass of Na2SO4 formed = Moles x MW Mol of H2S04 - Mass Mol of NaOH - Mass- 40.00g # Experimental molar ratio of reactants: 60.00 g MW 98.07984 gmol 0.61175 mol 0.50004 mol x 142.04314 g mol 71.027 g 1.00007 mol MW 39.997108 g mol # Step 3 Moles of H2SO4 consumed 0.5 x moles of NaOH consumed 0.5000 mol Mol of H2S04 Mol of NaOH 0.61175 mol_ 1.00007 mol # Remaining mol of H2SO4-0.61175-0.5000 # Remaining mass of H2SO4-0.11 171 mol x = 0.11 17 1 98.08 g mal" mol = 10.957 g # Comparing theoretical and experimental molar ratios of reactants, the experimental is less than its theoretical value 2.0 while keepring Mol of H2S04 constnat at Therefore Mol of NaOH 1.0 mol NaOH is the limiting reagent.