The equation is:
Ca(OH)₂(s) + 2 HCl(aq) → CaCl₂(aq) + 2 H₂
O(l)
n=mass in g/M.M
15 g Ca(OH)₂ is n=15 g/ 74.1 g/mol=0.2024 mol of Ca(OH)₂
no. of mol of HCl:
n=0.5 mol/L*0.075L=0.0375 mol
This could react with 0.0375/2= 0.01875 mol of Ca(OH)₂ We have a lot more than that.
Therefore, HCl is the limiting reagent and determines how much CaCl₂ forms.
Based on the balanced reaction, 2 moles of HCl gives 1 mole of CaCl₂
no. of mol of CaCl₂= 0.0375/2= 0.01875 mol
mass in g=n*MM= 0.01875*111= 2.08 g
