Please help me Part A, B, & C. It is very difficult and I tried to do it myself.

Question

Please help me Part A, B, & C. It is very difficult and I tried to do it myself. PLEASE HELP!!!!

Problem 9.50 Part A How many milliliters of 0.200M NaOHsolution are needed to react with 25.0 mL of a 0.470 MNiCl2 solution? Answer the following for the reaction NiCl2 (aq) + 2NaOH(aq) Ni(OH)2 (s) + 2NaCl(aq) Express your answer with the appropriate units alue Units Submit My Answers Give Up Incorrect; One attempt remaining; Try Again Part B How many grams of Ni (OH)2 are produced from the reaction of 52.0 mL of a 1.50 M NaOH solution and excess NiCl2? Express your answer with the appropriate units ValueUnits

Explanation / Answer

Look at the balanced chemical equation. As per the stoichiometric equation,

1 mole NiCl2 = 2 moles NaOH = 1 mole Ni(OH)2.

Part A

Millimoles of NiCl2 = (25.0 mL)*(0.470 M) = 11.75 mmole.

Millimoles of NaOH required to react completely with 11.75 mmole NiCl2 = (11.75 mmole NiCl2)*(2 mole NaOH/1 mole NiCl2) = 23.5 mmole NaOH.

Milliliters of 0.200 M NaOH required = (millimoles of NaOH required)/(molarity of NaOH) = (23.5 mmole)/(0.200 M) = 117.5 mL (ans).

Part B

We have excess NiCl2; therefore, NaOH is the limiting reactant. As per the stoichiometric equation,

2 moles NaOH = 1 mole Ni(OH)2

Millimoles of NaOH = (52.0 mL)*(0.150 M) = 7.8 mmole.

Millimoles of Ni(OH)2 produced = (7.8 mmole)*(1 mole Ni(OH)2/2 mole NaOH) = 3.9 mmole Ni(OH)2.

Molar mass of Ni(OH)2 = (1*58.6934 + 2*1.008 + 2*15.9994) g/mol = 92.7082 g/mol.

Mass of Ni(OH)2 produced = (3.9 mmole)*(1 mole/1000 mmole)*(92.7082 g/mol) = 0.36156 g 0.3616 g (ans).

Part C

We have already noted that 1 mole NiCl2 = 2 moles NaOH

Millimoles of NaOH = (17.8 mL)*(0.440 M) = 7.832 mmole.

Millimoles of NiCl2 reacted = (7.832 mmole NaOH)*(1 mole NiCl2/2 moles NaOH) = 3.916 mmole NiCl2.

Molarity of NiCl2 = (millimoles of NiCl2)/(volume of NiCl2 in mL) = (3.916 mmole)/(30.0 mL) = 0.1305 M (ans).

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