How many grams of nickel(II) chloride do you need to precipitate 503 mg of silver chloride in the reaction between nickel(II) ch

Question

2 years ago

13

loride and silver nitrate

1 answer:

maxonik [38] · Answer 1 · 2023-01-02T13:21:32+03:00

From the information provided in the question, the mass of NiCl2 required is 0.23 g.

NiCl2(aq) + 2AgNO3(aq) -------> 2AgCl(s) + Ni(NO3)2(aq)

Number of moles of AgCl= Mass/molar mass

Molar mass of AgCl= 143 g/mol

Substituting values;

Number of moles = 503 × 10^-3g/143 g/mol= 0.00352 moles

Since 1 mole NiCl2 precipitates 2 moles of AgCl

x moles of NiCl2 precipitates 0.00352 moles of AgCl

x = 1 mole × 0.00352 moles/2 moles

x = 0.00176 moles of NiCl2

Mass of NiCl2= 0.00176 moles × 130 g/mol = 0.23 g