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 excess Ca2+ will remain in excess.
C.Na2C2O4 is the limiting reagent.
D.Ca(NO3)2 is the excess reagent.
Solution
Ca(NO3)2 + Na2C2O4 → CaC2O4 ↓ + 2NaNO3
100 × 0.06 50 × 0.06
= 6 m mol = 3 m mol 3 m mol = 0.003 mol
Na2C2O4 is the limiting reagent.
∴ 3 m mol Na2C2O4 ≡ 3 m mol Ca(NO3)2
≡3 m mol CaC2O4 ≡ 6 m mol NaNO3
m mol of Ca(NO3)2 left = 6 - 3 = 3 m mol = 0.003 mol
MCa2+(left) =
= 0.02M
Hence, option (b) is wrong.
100 × 0.06 50 × 0.06
= 6 m mol = 3 m mol 3 m mol = 0.003 mol
Na2C2O4 is the limiting reagent.
∴ 3 m mol Na2C2O4 ≡ 3 m mol Ca(NO3)2
≡3 m mol CaC2O4 ≡ 6 m mol NaNO3
m mol of Ca(NO3)2 left = 6 - 3 = 3 m mol = 0.003 mol
MCa2+(left) =
= 0.02MHence, option (b) is wrong.
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