chem. check

Can u please check these questions..thanks!

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 one looks ok.

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 don't 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.

To three significant figures I obtained 66.5 g Zn required. also.

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.

I didn't answer #6 or #7 because they don't make sense to me. I will elucidate for each problem. Perhaps some other tutor knows more about this than I; if so, I invite them to share their expertise.

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 main problem with this is that I don't think there is such an animal as ferrous dichromate. (It isn't listed in the Chemical Rubber Handbook and it isn't listed in The Merck Index.)Even if it did exist in solid form, the dichromate would immediately oxidize the ferrous ion to ferric ion as soon as the solid was dissolved in water. Assuming such a stunt could be pulled off, then I wouldn't know which was in excess; i.e., will the dichromate oxidize the stannous ion to stannic ion and what happens to the ferrous ion. How much of the dichromate is used to oxidize the iron (that part we could calculate). At any rate, I added a phrase to the end of the final sentence as "Calculate the voluje of iron(II) dichromate required to do what? If you are simply looking at ferrous/ferric solution with Sn(II) to Sn(0) solution, then the answer I obtain is 0.890 M iron but it doesn't mean anything because of the above discussion.

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.

This is the same KIND of thing but a little different. First, calcium permanganate is a salt that is freely soluble in water; therefore, the salt will not precipitate from solution. Second, calcium metal reacts with water and it doesn't matter how much permanganate is there originally. Thus the limiting factor on how much calcium metal will react is the amount of water available and not the amount of permanganate available. Perhaps the problem is asking us to calculate how much calcium metal is required to form calcium permanganate IF we could prepare calcium permanganate with no excess calcium or calcium hydroxide present. If so, please clarify the question and post your work for me to check. I didn't find 73 or so grams. I found more like 11 g. I hope this helps.
My answer: 73.21g is the mass.

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