MasteringChemistry: Chapter 16 HW-Google Chrome Secure https: session masteringc

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MasteringChemistry: Chapter 16 HW-Google Chrome Secure https: session masteringchemistry.com myct itemView?assignment problemID-88223448 Chapter 16 HW Exercise 16.105 « previous | 11 of 53 next Exercise 16.105 Part A what volume o 0.118 M sodium carbonate solution s required to p ecipitate 99% of he Mg fr n 1.00 Express your answer to two significant figures and include the appropriate units of 0.100 M mag nitrate solution? V Value Units Submit My Answers give up Incorrect; Try Again; 3 attempts remaining Provide Feedback Continue 9-54 AM 11/10/2017 Type here to search

Explanation / Answer

Lets write the reaction equation first:

Na2CO3 + Mg(NO3)2 ----------> 2NaNO3 + MgCO3(s)

moles of magnesium nitrate = molarity x volume in liters = 0.100M x 1L = 0.1 moles

99% of 0.1 moles = 0.099 moles

1 mol Na2CO3 need 1 mol Mg(NO3)2 to react completely and form precipitate

0.099 moles of Mg(NO3)2 will need 0.099 moles Na2CO3

moles = molarity x volume in litres = 0.099 moles = 0.118 M x V

V = 0.83898 L ~ 0.84 L

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