Question

If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction is complete

Answer 1

Answer:

• 0.456M

Explanation:

1. Balanced molecular equation

     $2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O$

2. Mole ratio

     $\dfrac{2molHNO_3}{1molBa(OH)_2}$

3. Moles of HNO₃

• Number of moles = Molarity × Volume in liters
• n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

4. Moles Ba(OH)₂

• n = 0.700M × 0.0310 liter = 0.0217 mol

5. Limiting reactant

Actual ratio:

   $\dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28$

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

6. Final molarity of Ba(OH)₂

• Molarity = number of moles / volume in liters
• Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M