Question

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide. 4 HCl ( aq ) + MnO 2 ( s ) ⟶ MnCl 2 ( aq ) + 2 H 2 O ( l ) + Cl 2 ( g ) 4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g) A sample of 43.5 g MnO 2 43.5 g MnO2 is added to a solution containing 41.5 g HCl . 41.5 g HCl. What is the limiting reactant? MnO 2 MnO2 HCl HCl What is the theoretical yield of Cl 2 ? Cl2? theoretical yield: g Cl 2 g Cl2 If the yield of the reaction is 82.9 % , 82.9%, what is the actual yield of chlorine? actual yield: g Cl 2

Answer 1

Answer:

HCl is the limiting reactant.

20.2 grams = theoretical yield Cl2

actual yield = 16.75 grams Cl2

Explanation:

Step 1: Data given

Mass of MnO2 = 43.5 grams

Molar mass MnO2 = 86.94 g/mol

Mass of HCl = 41.5 grams

Molar mass HCl = 36.46 g/mol

Step 2: The balanced equation

4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles MnO2 = 43.5 grams / 86.94 g/mol

Moles MnO2 =  0.500 moles

Moles HCl = 41.5 grams / 36.46 g/mol

Moles HCl = 1.14 moles

Step 4: Calculate the limiting reactant

For 4 moles HCl we need 1 mol MnO2 to produce 1 mol MnCl2, 2 moles H2O and 1 mol Cl2

HCl is the limiting reactant. I will completely be consumed (1.14 moles).

MnO2 is in excess. There will react 1.14 /4 = 0.285 moles

There will remain 0.500 - 0.285 = 0.215 moles MnO2

Step 5: Calculate moles Cl2

For 4 moles HCl we need 1 mol MnO2 to produce 1 mol MnCl2, 2 moles H2O and 1 mol Cl2

For 1.14 moles HCl we'll have 1.14/4 = 0.285 moles Cl2

Step 6: Calculate mass Cl2

Mass Cl2 = moles Cl2 * molar mass Cl2

Mass Cl2 = 0.285 moles * 70.9 g/mol

Mass Cl2 = 20.2 grams = theoretical yield

Step 7: Calculate actual yield

% yield = (actual yield / theoretical yield) *100%

0.829 = actual yield / 20.2 grams

actual yield = 0.829 * 20.2 grams

actual yield = 16.75 grams Cl2