Chemistry help! givenn 2KMnO4 16 HCl(aq) ~~~~> 2MnCl2(aq) 5Cl2 (g) 8H2O(l) 2KCl(aq)..... how do I calc

Question

calculate mL of 6.0 M HCl required to react with 200 mL of .50 M KMnO4?

I know the answer is 13.0 mL but I keep getting the wrong answer. I know im doing the first part right just not the last part

200 ml KMnO4 x (.50 mol KMnO4/1000 mL KMnO4) = .1 mol KMnO4

.1 mol KMnO4 x (8HCl/1 KMnO4) = .0125 mol HCl

and then I dont know what to do?

Also will you help me do this I have my final tomorrow and I know this is going to be on the test!!!

(using the same balanced equation) calculate L of Cl2 gas at stp produced by the reaction of 75.0 mL of 6.0 M HCl

I know the answer is 15 L I just want to know the steps to get there becuase I will need this for my test and I cant google it tomorrow!

Answer 1

For the first part:

1) 0.2L KMnO4 X 0.5 mols/L = 0.1 mols  KMNO4

   for every 2 mols of KMNO4 it takes 16 Mols HCl

   therefore   0.1 mols KMNo4 X  16mols HCL /2 Mols KMNo4 = 0.8 mols of HCL

0.8 mols HCL X 1L HCL/6Mols = 0.133 L or 133 mls

2) 0.075 L HCL x 6 mols/L = 0.45 mols HCl

0.45 mols HCl X (5 mols Cl2 / 16 mols HCl) = 0.140 mols of Cl2 gas

each mol of gas at STP = 22.414L  

0.14 mols X (22.414L/mol) = 3.138 L

Answer 2

Moles KMnO4 = 0.200 L x 0.50 M =0.10

the ratio between KMnO4 and HCl is 2: 16

Moles HCl needed = 0.10 x 16 / 2 =0.80

V = moles / M = 0.80 / 6.0 =0.013 L =>  13 mL

Moles HCl = 0.0750 L x 6.0 M =0.45

the ratio between HCl and Cl2 is  16 : 5

Moles HCl = 0.45 x 5 / 16 =0.14

At STP 1 mole => 22.4 L

0.14 x 22.4 = 3.15 L

Answer 3

Moles KMnO4 = 0.200 L x 0.50 M =0.10

the ratio between KMnO4 and HCl is 2: 16

Moles HCl needed = 0.10 x 16 / 2 =0.80

V = moles / M = 0.80 / 6.0 =0.013 L => 13 mL

Moles HCl = 0.0750 L x 6.0 M =0.45

the ratio between HCl and Cl2 is 16 : 5

Moles HCl = 0.45 x 5 / 16 =0.14

At STP 1 mole => 22.4 L

0.14 x 22.4 = 3.15 L

Answer 4

2KMnO4 16 HCl(aq) ~~> 2MnCl2(aq) 5Cl2 (g) 8H2O(l) 2KCl(aq)

200 mL = 0,2 L

0,2 L KMnO4 x (0,50 mol KMnO4/L KMnO4) = 0,1 mol KMnO4

0,1 mol KMnO4 x (16HCl/2 KMnO4) = 0,8 mol HCl

0,8 mol HCl / (6,0 mol/L HCl) = 0,133 L HCl = 133 mL HCl      (Answer)

Answer 5

you have 0.1mol of KMnO4. you need 8 times more HCL so that is 0.8mols. 0.8mols/6M gives 0.133 litres or 133mL. you were out by a factor of 10.

secondly the number of moles of HCL used is 0.45mols ((6M/1000)*75) and from that the number of moles of CL2 is 0.140mols((0.45mols/16)*5). according to http://hyperphysics.phy-astr.gsu.edu/hba...

one mole of gas at STP occupies 22.4litres so 0.140*22.4 is 3.136 litres.

i hope this helps and good luck with the test tomorrow.