Use the equation below to calculate the following:2 NaOH(aq) + H2SO4(aq)-->Na2SO4(aq) + 2 H2O(l) Assignment > Open AssignmentPRINTER VERSION1 BACKNEXTIASSIGNMENTRESOURCESHein Paired Exercise 14.36Use the equation below to calculate the following:Chapter 14HomeworkM Hein PracticeExercise 14.12 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H20(1)M Hein PracticeExercise 14.3(a) the moles Na2S04 produced from 4.7 mol H2SO4E Hein PracticeExercise 14.4M Hein PracticeExercise 14.5mol Na2SO4E Hein PracticeExercise 14.6M Hein PracticeExercise 14.7the tolerance is +/-1 in the 2nd significant digitE Hein PracticeExercise 14.8(b) the moles H20 produced from 0.041 mol NaOHHein PracticeExercise 14.9mol H20M Hein PracticeExercise 14.10M Hein Paired Exercise14.2E Hein Paired Exercisethe tolerance is +/-1 in the 2nd significant digit14.4Hein Paired Exercise14.36(c) the moles NaOH required to react with 1.60 L of 0.167 M H2SO4O Hein ReviewQuestion 14.16M Hein ReviewQuestion 14.8mol NaOHthe tolerance is +/-1 in the 3rd significant digit Assignment > Open AssignmentPRINTER VERSION(BACKNEXTASSIGNMENTRESOURCES(d) the grams of Na2S04 that can be obtained from 26 mL of 0.057 M NaOHChapter 14HomeworkM Hein PracticeExercise 14.1M Hein PracticeExercise 14.3|g Na2S04the tolerance is +/-1 in the 2nd significant digitM Hein PracticeExercise 14.4M Hein PracticeExercise 14.5(e) the volume of 0.259 M H,SO4 needed to react with 33.7 mL of 0.777 M NaOHM Hein PracticeExercise 14.6M Hein PracticeExercise 14.7M Hein PracticeExercise 14.8ml H2SO4the tolerance is +/-1 in the 3rd significant digitHein PracticeExercise 14.9M Hein PracticeExercise 14.10(f) the molarity (M) of the NaOH solution when 51.76 mL react with 55.57 mL of 0.197 M H2SO4O Hein Paired Exercise14.2M Hein Paired ExerciseM NaOH14.4the tolerance is +/-1 in the 3rd significant digitHein Paired Exercise14.36O Hein ReviewQuestion 14.16M Hein ReviewQuestion 14.8SHOW HINT

Since you have posted question with multiple sub-parts, we are entitled to answer the first 3 only. The balanced reaction given is, => 2 NaOH (aq) + H2SO4 (aq) --------> Na2SO4 (aq) + 2 H2O (l) a) Given : moles of H2SO4 reacting = 4.7 mol From the above balanced reaction we can see that 1 mole of H2SO4 is reacting to produce 1 mole of Na2SO4. Hence moles of Na2SO4 produced = moles of H2SO4 reacting = 4.7 mol Hence the answer is 4.7 mol Na2SO4. b) Given : moles of NaOH reacting = 0.041 mol From the above balanced reaction we can see that 2 moles of NaOH is reacting to produce 2 moles of H2O. Hence moles of H2O produced = moles of NaOH reacting = 0.041 mol Hence the answer is 0.041 mol H2O.c) Given : concentration of H2SO4 = 0.167 M Volume of H2SO4 solution reacting = 1.60 L Since moles = concentration X volume of solution in L => Moles of H2SO4 reacting = 0.167 X 1.60 = 0.2672 mol approx. From the above balanced reaction we can see that 2 moles of NaoH is reacting with 1 mole of H2SO4 Hence moles of NaOH required = moles of H2SO4 reacting X 2 = 0.2672 X 2 = 0.5344 mol Hence the answer is 0.534 mol NaOH approx (in 3 significant digits).