How many grams of solid calcium hydroxide, Ca(OH)2, are required to react with 350 mL of 0.40 M HCl? __HCl + __Ca(OH)2  __CaCl2 + __H2O

Oct 18, 2014

2HCl (aq) + Ca${\left(O H\right)}_{2}$ (aq) ---------> Ca$C {l}_{2}$ (ppt) + 2${H}_{2}$O (aq)

let us calculate the number of moles , as per the chemical reactions;

2 moles of HCl solution reacts with one mole Calcium Hydroxide
Ca${\left(O H\right)}_{2}$
One mole of HCl has mass : 36.5 g/mol, two moles of HCl will have mass, 73 g.

One mole of Ca${\left(O H\right)}_{2}$ has mass 74.1 g

as per equation; 73 g of HCl reacts with 74.1 g of Ca${\left(O H\right)}_{2}$

1g of HCl reacts with 74.1g / 73 of Ca${\left(O H\right)}_{2}$

1g of HCl reacts with 1.015 g of Ca${\left(O H\right)}_{2}$

NOW AS PER THE QUESTION MOLARITY AND VOLUME OF HYDROCHLORIC ACID IS GIVEN, IT CAN BE USED TO CALCULATE THE MASS OF HYDROCHLORIC ACID IN THE SOLUTION.

NUMBER OF MOLES of HCl ; Molarity of solution x Volume of Solution

 of moles of HCl = (0.40 mol/L ) x 350 mL

= (0.40 mol/L ) x 0.350 L = 0.14 mol

mass of HCl that makes 0.14 mol of HCl =  of moles x molar mass of HCl

mass of HCl = 0.14 mol x 36.5 g/ mol

mass of HCl = 5.11g

As per Stoichiometry , 1g of HCl reacts with 1.015 g of Ca${\left(O H\right)}_{2}$, 5.11g of HCl can react with 5.11 x 1.015 = 5.1865 g or 5.2 g of
Ca${\left(O H\right)}_{2}$ .