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

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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

Learn more: https://brainly.com/question/2510654

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