Can I get this broken down please? Chlorine gas can be prepared in the laborator

Question

Can I get this broken down please?

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide: 4HCl(aq)+MnO2(s)-->MnCl2(aq)+2h2O(l)+Cl2(g) you add 34.5g of MnO2 to a solution containing 41.5g of HCl. What is the limiting reactant? What is the Theoretical yield of Cl2? If the yield of the reaction is 80.5%, what is the actual yield of chlorine?

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide You add 34,5 g of Mno2 to a solution containing 41.5 g of HCI. (a) What is the limiting reactant? O Mno2 o HCI (b) What is the theoretical yield of Cl2? Number Incorrect .284 g CI, You have given the moles of chlorine formed. Convert to grams. (c) If the yield of the reaction is 80.5%, what is the actual yield of chlorine? Number 22.862. g Cl,

Explanation / Answer

a)

mol of MnO4 = mass/MW = 34.5/86.9368 = 0.396840 mol

mol of HCL = mass/MW = 41.5/36.5 = 1.1369 mol

therefore, ratio is 1:4

0.396840 mol ofMnO4 = 4*0.396840 = 1.58736 mol of HCl required, which we do NOT have

so Hcl is limiting

b)

theoretical yield of Cl2

4 mol of HCl = 1 mol of Cl2

1.58736 mol of HCl = 1/4*1.58736 = 0.39684 mol of Cl2

mass = mol*MW = 0.39684*70.89 = 28.131 g of Cl2 will be produced

c)

if yield is 80.5%

then, actual yield = theoretical * %yield / 100 %= 28.131 * 80.5/100 = 22.645 g of Cl2 expected

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