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#1
Jan 27 2015 07:05pm
A 78.4 CM^3 of HCl solution requires 1.240 g of pure CaCO3 for neutralization. Calculate the normality of the acid.
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#2
Jan 27 2015 11:35pm
each co3^2- can neutralize 2 H+s.

caco3= 100g/mol
= 0.00124mol

thus there are 0.00248mol of H+ in solution

78.4 CM^3= 0.0784 L

HCL has equivalence of 1

0.00248/0.0784
=0.0316N
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