Calculate the volume, in mL, of 1.72 M HCl solution that will react with 2.67 mol of CaCO3?
HCI + CaCO3 ---> CaCl2 + H2O + CO2
C(HCl) = 1.72 M;
n(CaCO3) = 2.67 mol;
2HCI + CaCO3 ---> CaCl2 + H2O + CO2;
For following chemical reaction:
n(HCl) = 2 * n(CaCO3) = 2 * 2.67 = 5.34 mol;
C(HCl) = n(HCl)/V(HCl);
V(HCl) = n(HCl)/C(HCl) = 5.34/1.72 = 3.105 L = 3105 mL.
Answer: 3105 mL.
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