Question

what weight of CaCO3 shall be neutralised by 100 ml of 0.1 M HCl?

Answer 1

A properly balanced equation for the reaction is –

CaCO₃+ 2HCl → CaCl₂ + CO₂ + H₂O i.e 1 mol CaCO₃ will react with 2 mol HCl 

100 ml of 0.1 M HCl is used. Therefore, 100 / 1000 x 0.1 = 1 mol HCl

 

According to the equation, 1/ 2 = 0.5 mol CaCO₃

Therefore, 0.5 mol CaCO₃ is neutralized by 100 ml of 0.1 M HCl.

Answer 2

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