Answered

Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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

Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.