Asked by cheri
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
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
Answers
Answered by
DrBob222
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
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