V NaOH = 22,6 cm³ = 0,0226dm³
c mol NaOH = 0,102 mol/dm³
c mol = n / V
n NaOH = 0,102 mol/dm³ * 0,0226dm³ = 23 * 10⁻⁴ mol
HCl + NaOH ---> NaCl + H₂O
1 mol HCl reaguje z 1 molem NaOH
n HCl = n NaOH = 23 * 10⁻⁴ mol <---- nadmiar kwasu
-------------------------------------------------------------------------
m CaCO₃ = 0,265 g
M CaCO₃ = 100 g/mol
CaCO₃ + 2HCl ---> CaCl₂ + CO₂ + H₂O
1 mol CaCO₃ reaguje z 2 molami HCl
100 g CaCO₃ ------------- 2 mole HCl
0,265 g CaCO₃ ---------- x moli HCl
n HCl = 0,265 g * 2 mol / 100 g
n HCl= 53 * 10⁻⁴ mol <--- kwas, który przereagował
------------------------------------------------------------------------------
∑ n HCl = 23 * 10⁻⁴ mol + 53 * 10⁻⁴ mol = 76 * 10⁻⁴ mol
c HCl = 0,152 mol/dm³
V HCl = 76 * 10⁻⁴ mol / 0,152 mol/dm³ = 0,05 dm³ = 50 cm³
