d0r u.0T NaCI 0.0500 mole of Ca(NO;)2 Iton to obtain 3.00 mo b. liters of a 1.50
ID: 1033377 • Letter: D
Question
d0r u.0T NaCI 0.0500 mole of Ca(NO;)2 Iton to obtain 3.00 mo b. liters of a 1.50 M NaCl solution to obtain 15.0 mo c. milliliters of a 0.800 M Ca(NO3)2 solution to obtain For each of the following solutions, calculate the: a. liters of a 4.00 M KCI solution to obtain 0.100 mole of K b. liters of a 6.00 M HCI solution to obtain 5.00 moles of HO c. milliliters of a 2.50 M K2SO4 solution to obtain 1.20 moles 18 of K,so, 19 Answer the following for the reaction: Pb(NO3)2(aq) +2KCI(aq) PbCl2(s) +2KNO;(aq) a. How many grams of PbCl2 will be formed from 50.0 mL of b. How many milliliters of a 2.00 M Pb(NO3)2 solution will c. What is the molarity of 20.0 mL of a KCI solution that a 1.50 M KCl solution? react with 50.0 mL of a 1.50 M KCI solution? reacts completely with 30.0 mL of a 0.400 M Pb(NO3)2 solution? 50 Answer the following for the reaction: NiCl2(aq) + 2NaOH(aq)--> Ni (OH)2(s) +2NaCl(a?) a. How many milliliters of a 0.200 M NaOH solution are needed to react with 18.0 mL of a 0.500 M NiCl2 solution? b. How many grams of Ni (OH)2 are produced from the reac- tion of 35.0 mL of a 1.75 M NaOH solution and excess NiCI2?Explanation / Answer
(50)
Moles of NiCl2 = Molarity*Volume = 0.5*0.018 = 0.009
As per the reaction stoichiometry, moles of NaOH reqd = 2*0.009 = 0.018
So,
Volume of NaOH solution required = Moles/Molarity = 0.018/0.2 = 0.09 L = 90 mL
(b)
Moles of NaOH taken = 1.75*0.035 = 0.06125
So,
Moles of Ni(OH)2 produced = 0.06125/2 = 0.030625
Mass of Ni(OH)2 produced = Moles*MW = 0.030625*92.7 = 2.84 g
Hope this helps !
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