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
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