a, n HCl =44,8/22,4 = 2 mol
m HCl = 2 x 36,5 =73g
C% dd A = 73/(73+327) x 100% = 18,25%
b, mHCltrong 250g ddA = 250x18.25/100 = 45.625 (g)
nHCltrong 250g ddA = 45.625/36.5 = 1.25 (mol)
nCaCO3 = 50/100 = 0.5 (mol)
Pư:
CaCO3 + 2HCL ---> CaCl2 + H2O + CO2
_0.5____[1.25]_____0.5_________0.5_
nHCldư = 1.25 - 0.5x2 = 0.25 (mol)
mHCldư = 0.25 x 36.5 = 9.125 (g)
mddB = 50 + 250 - 0.5x44 = 278 (g)
C%CaCl2 = (0.5x111x100) / 278 = 19.964%
C%HCldư = (9.125x100) / 278 = 3.282%
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