Question

n the reaction NiCl2(aq) + 2NaOH(aq) → Ni(OH)2(s) + 2NaCl(aq) How many milliters (mL) of 0.836 M NaOH are needed to react with 12.8 mL of 0.500 M NiCl2?

Answer 1

Number of moles of NiCl2 = molarity * volume of solution in L

Number of moles of NiCl2 = 0.500 * 0.0128 = 0.0064 mole

From the balanced equation we can say that

1 mole of NiCl2 requires 2 mole of NaOH so

0.0064 mole of NiCl2 will require

= 0.0064 mole of NiCl2 *(2 mole of NaOH / 1 mole of NiCl2 )

= 0.0128 mole of NaOH

Molarity of NaOH = number of moles of NaOH / volume of solution in L

0.836 = 0.0128 / volume of solution in L

volume of solution in L = 0.0128 / 0.836 = 0.0153 L

1L = 1000 mL

0.0153 L = 15.3 mL

Therefore, the volume of NaOH needed would be 15.3 milliliter