One gram mixture of caco3 and nacl react completely with hundred ml of invite in n/10 hcl the percentage of caco3 in the mixture is
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Answer:
Explanation:
=> We can write the chemical reaction as follow for the given question:
CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O
=> 1 mol CaCO₃ = 2 mol HCl
moles of HCl = 100 mL * 0.1 N
= 100 / 1000 * 0.1
= 0.01 moles
=> Moles of CaCO₃ = 0.01 / 2 = 0.005 moles
Mass of CaCO₃ = 0.005 mol * 100 g/ml
= 0.5 gram
=> Now, in 1 gm of mixture of NaCl and CaCO₃
mass of NaCl = 1 - 0.5 = 0.5 g
% of NaCl = 0.5 g / 1 g * 100
= 50%
Thus, the percentage of caco3 in the mixture is 50%
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