Mole ConceptHard
Question
100 mL of 0.06 M Ca(NO3)2 is added to 50 mL of 0.06 M Na2C2O4. After the reaction is complete.
Options
A.0.003 moles of calcium oxalate will get precipitated
B.0.003 M of Ca2+ will remain in excess
C.Na2C2O4 is limiting reagent.
D.Ca(NO3)2 is excess reagent.
Solution
(A), (C) and (D) Explanation :
Ca2+ + C2O42- → CaC2O4
100 ml of 0.06 M Ca(NO3)2 contains 100 × 0.06 = 6 millimole
50 ml of 0.06 M Na2C2O4 contains 50× 0.06 = 3 millimole
3 millimoles of C2O42- will react with 3 millimoles of Ca2+ for 3 millimoles of CaC2O4
=
= 0.003 moles
(A) is correct.
Excess of Ca2+ left = 6 - 3 = 3 millimoles =
= 0.003 moles.
(C) is correct because sodium oxalate is present in smaller amount than required.
(D) is correct because Ca(NO3)2 is present in large amount.
Ca2+ + C2O42- → CaC2O4
100 ml of 0.06 M Ca(NO3)2 contains 100 × 0.06 = 6 millimole
50 ml of 0.06 M Na2C2O4 contains 50× 0.06 = 3 millimole
3 millimoles of C2O42- will react with 3 millimoles of Ca2+ for 3 millimoles of CaC2O4
=
= 0.003 moles(A) is correct.
Excess of Ca2+ left = 6 - 3 = 3 millimoles =
= 0.003 moles.(C) is correct because sodium oxalate is present in smaller amount than required.
(D) is correct because Ca(NO3)2 is present in large amount.
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