Answered

Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

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

Sagot :

pstnonsonjoku

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

We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.