Wednesday

July 1, 2015

July 1, 2015

Posted by **Dave** on Wednesday, July 12, 2006 at 4:45pm.

Redox Stoichiometry

1.A 2.45g zinc metal strip is placed into an 2.3 x 10-1 mol/L solution of tin (II) bromide. If all the zinc and tin (II) bromide react, what was the volume of the solution.

My answer: 8.62L is the volume of the

solution.

3.624 mL of potassium permanganate is titrated with 550 mL of an acidic solution of 0.350 mol/L iron (II) chloride. What was the concentration of the potassium permanganate?

My answer: 0.62mol/L is the concentration.

4.A 2.34 mol/L sodium hydroxide solution is reacted with 11.6 g of lead (II) sulphate powder. What volume of sodium hydroxide would react completely?

My answer: 0.033L is the volume.

6.A 4.90 mol/L solution of iron (II) dichromate is combined with 3.50L of a

1.25 mol/L acidic solution of tin (II) bromide. Calculate the volume of iron (II) dichromate required.

My answer: 0.204L is the volume.

7. If calcium metal is added to 86.9 g of potassium permanganate in an acidified solution until the reaction is complete. Calculate the mass of calcium metal that would have reacted.

My answer: 73.21g is the mass.

10.If 278 mL of a 1.22 mol/L acidic solution of ammonium dichromate reacts with a zinc metal strip, what mass of zinc will react?

My answer: 66.53 is the mass.

Thanks!

1.A 2.45g zinc metal strip is placed into an 2.3 x 10-1 mol/L solution of tin (II) bromide. If all the zinc and tin (II) bromide react, what was the volume of the solution.

Hi.

This is my work for this one:

Zinc moles (n):

(2.45g/65.38g/mol)=0.03747moles

ratio of equation:

(1/1)

Molar mass of Tin(2)bromide:

(2.3x10^-1mol/L)

So now, just multiply:

=(0.03747moles)(1/1)(2.3x10^-1mol/L)

=0.0086L

I may have just forgot to punch it in right. So would this be right?

If molarity = mols/L, then

L = mols/M = 0.03747 mols/0.23 molar = ??

I hope this helps.

3.624 mL of potassium permanganate is titrated with 550 mL of an acidic solution of 0.350 mol/L iron (II) chloride. What was the concentration of the potassium permanganate?

My answer: 0.62mol/L is the concentration.

4.A 2.34 mol/L sodium hydroxide solution is reacted with 11.6 g of lead (II) sulphate powder. What volume of sodium hydroxide would react completely?

My answer: 0.033L is the volume

This is what I did:

#3.iron (II) chloride(n):

(0.350mol/L x0.550L)= 0.1925moles

ratio of equation:

(1/5)

Molar mass of Tin(2)bromide:

(1/0.624L)

So now, just multiply:

=(0.1925moles)(1/5)(1/0.624L)

=0.06169877179L

Rounded: 0.062mol/L

Can you please help me, because I dont know what I am doing wrong. Thanks so much!

This is very good. Look at your original post and you will see what "I misread it as" because the problem #3 has no space between it and the 0.624 mL; therefore, I read the volume as 3.624 mL and came out with an outlandish 10+ for molarity. You could avoid errors of this type by always placing a zero in front of a decimal, such as 0.624 mL. Also, placing the number of the problem as (#3) helps. Your work is good. Finally, posting just one problem per post helps you get an answer faster because some tutors don't have the time to check all of them at one time. I hope this helps.

10.If 278 mL of a 1.22 mol/L acidic solution of ammonium dichromate reacts with a zinc metal strip, what mass of zinc will react?

My answer: 66.53 is the mass.

Hi,I was just wondering if the answer was right and if you could check #6 and #7. Thanks for your help!

I thought I answered this. I obtained an answer of 66.51g vs your 66.53g but since there are only three significant figures in 1.22M and 278 mL, then we are allowed only three in the final answer; therefore, 66.5g Zn is the correct answer but I wouldn't think 66.51 would be counted incorrect.

6.A 4.90 mol/L solution of iron (II) dichromate is combined with 3.50L of a

1.25 mol/L acidic solution of tin (II) bromide. Calculate the volume of iron (II) dichromate required.

My answer: 0.204L is the volume.

7. If calcium metal is added to 86.9 g of potassium permanganate in an acidified solution until the reaction is complete. Calculate the mass of calcium metal that would have reacted.

My answer: 73.21g is the mass.