Chem 高手請答~~

Question

1.The following experiment was conducted to determine the % by mass of calcium carbonate in an impure sample.

Step1 1.05 of the sample were weighted.
Step2 25 cm^3 of 0.86M nitric acid were added to react with the sample.
Step3 The excess acid was titrated against 0.6 M sodium hydroxide sol. .
22cm^3 of the alkali were required to reach the end point.

ans: 79.1%
我計到39.6%
plz explain thz~

Answer 1

Reaction in step 2 :
CaCO3 + 2HNO3­ → Ca(NO3)2 + H2O + CO2 .... (*)
Reaction in step 3 :
NaOH + HNO3 → NaNO3 + H2O .... (**)

Consider reaction (**).
Mole ratio NaOH : HNO3 = 1 : 1
No. of moles of NaOH used = 0.6 x (22/1000) = 0.0132 mol
No. of moles of HNO3 used in (**) = 0.0132 mol

Consider reaction (*).
Total number of moles of HNO3 used = 0.86 x (25/1000) = 0.0215 mol
No. of moles of HNO3 used in (*) = 0.0215 - 0.0132 = 0.0083 mol
Mole ration CaCO3 : HNO3 = 1 : 2
No. of moles of CaCO3 = 0.0083/2 = 0.00415 mol
Molar mass of CaCO3 = 40.1 + 12 + 16x3 = 100.1 g mol-1
Mass of CaCO3­ in the sample = 0.00415 x 100.1 = 0.415 g

% by mass of CaCO3 = (0.415/1.05) x 100% = 39.5%

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The correct answer is 39.5% or 39.6%.
If the mole ratio in reaction (*) is incorrectly taken as 1 : 1, the corresponding incorrect answer would be 79.0% or 79.1%.