Calculate the% by mass of calcium carbonate in the sample

[匿名]

2008-11-10 11:38 pm

step1
11g of sample were placed in a beaker
step 2
120cm^3 of 2M nitric acid were added to react with the sample
step 3
the excess acid was titrated against 1.60M sodium hydroxide solution
with phenolphthalein as indicator. 17.5cm^3 of the alkali were
required to reach the endpoint.
R.A.M:Ca=40.1,C=12.0,O=16.0

回答 (1)

Uncle Michael

2008-11-11 2:17 am

✔ 最佳答案

Consider the titration of excess HNO3 (unreacted with the sample) against NaOH.

HNO3 + NaOH → NaNO3 + H2O

Mole ratio HNO3 : NaOH = 1 : 1

No. of moles of NaOH used = 1.6 x (17.5/1000) = 0.028 mol

No. of moles of HNO3 unreacted with the sample = 0.028 mol

Consider the reaction of HNO3 with the sample of impure CaCO3:

(Assume the impurities do not cause any reaction.)

CaCO3 + 2HNO3 → Ca(NO3)2 + H2O + CO2

Mole ratio CaCO3 : HNO3 = 1 : 2

Total number of HNO3 added = 2 x (120/1000) = 0.24 mol

No. of moles of HNO3 unreacted = 0.028 mol

No. of moles of HNO3 reacted = 0.24 - 0.028 = 0.212 mol

No. of moles of CaCO3 used = 0.212 x (1/2) = 0.106 mol

Molar mass of CaCO3 = 40 + 12 + 16x3 = 100 g mol-1

Mass of CaCO3 used = 0.106 x 100 = 1.06 g

% by mass of CaCO3 in the sample = (1.06/1.1) x 100% = 96.36%
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