If 310mL of 2.2molar HCl solution is added to
135 mL of 3.6 molar Ba(OH)2 solution, what
will be the molarity of BaCl2 in the resulting
solution?
Answer in units of M
2HCl + Ba(OH)2 ==> 2H2O + BaCl2
mols HCl = M x LO = 0.310 x 2.2M = about 0.7 but you do it more accurately.
mols Ba(OH)2 = 0.135 x 3.6 = approximately 0.5.
Which is the limiting reagent? Convert mols HCl and mols Ba(OH)2 to mols BaCl2.
mols BaCl2 = 0.7 mol HCl x (1 mol BaCl2/2 mol HCl) = 0.7 x (1/2) = 0.35 mol BaCl2.
mols BaCl2 = 0.5 x [1 mol BaCl2/1 mol Ba(OH)2] = 0.5 x 1/1 = 0.5 mols BaCl2.
In limiting reagent problems it's the smaller value of product that ALWAYS wins; therefore, 0.35 mol BaCl2 will be formed. M = mols/L soln.
mols = 0.35
L = 0.310 + 0.135 = ?L