Question

Show all of your work for full credit and report your final answer with the correct number of significant figures and units. Solid MnO2 and HCl in an aqueous solution react to form MnCl2 in an aqueous solution, gaseous Clz and liquid H:O. a) (5 points) Write and balance the chemical equation. Be sure to show the phases of the reactants and products. b) (20 points) 20.0 g of MnO2 and 300 g of HCI react. Which is the limiting reactant? Show all of your work and explain your reasoning for full credit S20 CHEM 111-03 & 04 Take-Home Quiz 7 c) (10 points) What is the theoretical yield of MnCl? Show all of your work and explain your reasoning for full credit

Answer 1

D. a) Mno, t 4 HCl → Mrce + 2H₂O + c Given, 20.0g 30.0g of of MnO₂ the Find u moles with limiting of He 1 mole reactant is required of Mnog to react mol MnO₂ = grams Mno₂ g/mol MnO₂ 20.0 86.9 0.23 moles mol HCl = grams Hce g/mol HCl 30.0 36.46 0.82 moles = required a te MnO : 1 mol Амок = 1 4 = 0.25 mol of Но, actual : mol HCl = 0.82 = 3.56 mol of ol Mno, O, 23 н. у since the actual ratio is larger than required, He is the exess reagent, Mroz is L.R.

- moles Moles a uco of Hce = = 0.82 moles 30.0 36.46 = 20.0 86.9 - moles of mno, = 0.23 moles Since, MnO which is we need to Mice will is the limiting reactant producing Mrch therefore determine how many moles of be produced for Mall , theoretical field of .- 6 and do 1 mol Mno, - → 1 mal Mach molar mass of Macl = 125.844 g/mol since 0.23 mol of MnO₂ - so mo of Mrich = 0.23 = 0.23 mass of yield) Mrch (theoretical = 0.23 X 125.844 = 28.944.

dy Assuming all of MnO2 is used up, -2344 0.0575 mol of try the 0.23 x 4 = 0.92 mol of Hce req therefore remaining He a = 0.10 moles _