1) how many miligrans of potassium dichromate heptahydrate are needed to prepare

Question

1) how many miligrans of potassium dichromate heptahydrate are needed to prepare 50 mL of 0.250M aqueous solution that is 2.87 M? describe the procedure for making this solution.

2) to wwhat volume should 200 mL of a 15.0 M solution of sodium hydroxide be diluted to obtain a solution that is 2.87 M?

3) 35.00 mL of 0.25 M copper (ii) nitrate is mixted with 27.25 mL of 0.325 M aluminum (ii) carbonate resulting in a copper (ii) carbonate precipitate. Write a balanced chemical reaction, determine the limiting reagent and calculate the grams and moles of the precipitate.

Explanation / Answer

1) given

50 ml of 0.25 M

mass of solute = molarity x volume (L) x molar mass

mass of solute = 0.25 x 0.05 x 420

mass of solute = 5.25

so 5.25 g of potassium dichromate heptahydrate is required

intially it is 2.87 M , finally it is 50 ml and 0.25 M

so

apply M1V1 = M2V2

2.87 x V1 = 0.25 x 50

V1 = 4.355 ml

so initial volume is 4.355 ml

so take 4.355 ml of 2.87 M solution of potassium dichromate heptahydrate

then dilute it to 50 ml with water to get a solution of 0.25 M
2)

moles = molarity x volume

while diltuion moles remain same

so

intial moles = final moles

M1V1 = M2V2

15 x 200 = 2.87 x V2

V2 = 1045 ml

V2 = 1.045 L

so

the solution should be diluted to 1.045 L

3)

moles = molarity x volume (L)

moles of Cu(N03)2 = 0.25 x 0.035

moles of Cu(N03)2 = 8.75 x 10-3

moles of Al2(C03)3 = 0.325 x 0.02725

moles of Al2(C03)3 = 8.85625 x 10-3

3Cu(N03)2 + Al2(C03)3 ---> 3CuC03 + 2Al(N03)3

from the above reaction

moles of Al2(C03)3 required = moles of Cu(N03)2 / 3

moles of Al2(C03)3 required = 8.75 x 10-3 / 3

moles of Al2(C03)3 required = 2.9167 x 10-3

but 8.85625 x 10-3 moles of Al2(C03)3 is present. it is in excess

so Cu(N03)2 is the limiting reagent

from the reaction

moles of CuC03 formed = moles of Cu(N03)2 reacted = 8.75 x 10-3

so

moles of precipitate = 8.75 x 10-3

mass = moles x molar mass

mass of CUC03 = 8.75 x 10-3 x 123.55

mass of CuC03 = 1.08 g

so 1.08 g of precipitate is formed

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.