Sorry.my another question is wrong.here is my question?

'B' beaker's solution is equivalent to 'A' beaker's solution.enter image source here

1 Answer
Dec 3, 2017

Warning! Long Answer. The solution in "B" is not completely oxidized by the solution in "C".

Explanation:

Step 1. Calculate the moles of #"Cr"_2"O"_7^"2-"# in Beaker A

#"Moles of Cr"_2"O"_7^"2-" = 0.148 color(red)(cancel(color(black)("L Cr"_2"O"_7^"2-"))) × ("0.01 mol Cr"_2"O"_7^"2-")/(1 color(red)(cancel(color(black)("L Cr"_2"O"_7^"2-")))) = "0.001 48 mol Cr"_2"O"_7^"2-"#

Step 2. Calculate the moles of #"S"_2"O"_3^"2-"# in Beaker B

The stoichiometric equation for a reaction between dichromate and thiosulfate is

#"Cr"_2"O"_7^"2-" + 6"S"_2"O"_3^"2-" + 14"H"^"+" → "2Cr"^"3+" +"3S"_4"O"_6^"2-" +7"H"_2"O"#

#"Moles of S"_2"O"_3^"2-" = "0.001 48" color(red)(cancel(color(black)("mol Cr"_2"O"_7^"2-"))) × ("6 mol S"_2"O"_3^"2-")/(1 color(red)(cancel(color(black)("mol Cr"_2"O"_7^"2-")))) = "0.008 88 mol S"_2"O"_3^"2-"#

3. Calculate the moles of #"I"_2# formed in Beaker C

The #"I"^"-"# in Beaker C will react with the #"Cu"^"2+"# to form #"I"_2#.

#"2Cu"^"2+" + "2I"^"-" → "2Cu"^"+" + "I"_2#

If the #"KI"# is in excess, the #"CuSO"_4# will be the limiting reactant.

#"Moles of Cu"^"2+"#

#= 2.0 color(red)(cancel(color(black)("g CuSO"_4·5"H"_2"O"))) × (1 color(red)(cancel(color(black)("mol CuSO"_4·5"H"_2"O"))))/( 249.68 color(red)(cancel(color(black)("g CuSO"_4·5"H"_2"O")))) × "1 mol Cu"^"2+"/(1 color(red)(cancel(color(black)("mol CuSO"_4·5"H"_2"O")))) = "0.008 01 mol Cu"^"2+"#

#"Moles of I"_2color(white)(l) "formed" = "0.008 01" color(red)(cancel(color(black)("mol Cu"^"2+"))) × "1 mol I"_2/(2 color(red)(cancel(color(black)("mol Cu"^"2+")))) = "0.004 00 mol I"_2#

Step 4. Calculate the moles of iodine needed to react with Beaker B

The equation for the reaction is

#2"S"_2"O"_3^"2-" + "I"_2 → "2I"^"-" + "S"_4"O"_6^"2-"#

#"Moles of I"_2 = "0.008 88" color(red)(cancel(color(black)("mol S"_2"O"_3^"2-"))) × "1 mol I"_2/(2 color(red)(cancel(color(black)("mol S"_2"O"_3^"2-"))))= "0.004 44 mol I"_2#

Step 5. Was there enough iodine in Beaker C to completely oxidize the contents of beaker C?

Beaker C contained 0.004 00 mol #"I"_2#.

Beaker B required 0.004 44 mol #"I"_2#.

Thus Beaker C did not contain enough iodine to completely oxidize the contents of Beaker B

The contents of Beaker B were not completely oxidized by the contents of Beaker C.