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Secure! https://bbcsueastbay.edu/ultralcoursesL40984.Veyoutine Question Completion Status: A 3.2478 g sample of an unknown mixture of salts containing chloride produced 11.7154 g of AgCis) when reacted with excess AgNO3 What is the percentage (by mass) of choride in the original unknown chloride? (Answer to two decimal places QUESTION 14 Consider the following precipitation reaction 2 Na3PO4(aq) + 3 CuCl2(aq) CuaPolz(s) . 6 Nac (aq) What volume of 0.239 M Na3PO4 solution is necessary to completely react with 59.9 mL of 0.110M CuCl2? QUESTION 15 How many g of Agci can be produced by 32.1 ml of 0.256 M NaCl and an excess of 0.250 M AgNO3 solution? Save Al and Submait to save and submit. Click Save All Ansors to se all nsurs

Explanation / Answer

1 mole of AgCl contains 1 mole of Cl^-
143.5g of AgCl constins 35.5g of Cl^-
11.7154g of AgCl contains = 35.5*11.7154/143.5 = 2.9g of Cl^-

the percentage of Cl^- in unknown sample = 2.9*100/3.2478 = 89.3%
Q74
    2Na3Po4 + 3CuCl2 -------> Cu3(PO4)2 + 6NaCl
    2 moles    3 moles
   Na3PO4                        CuCl2
   M1 = 0.239M                   M2 = 0.11M
   V1   =                         V2 = 59.9ml
   n1 = 2                        n2 = 3
           M1V1/n1   =   M2V2/n2
              V1     =   M2V2n1/M1n2
                     =   0.11*59.9*2/0.239*3   = 18.4ml>>>>answer
Q15.
    NaCl + AgNo3 --------> NaNO3 + AgCl
   no of moles of NaCl = molarity * volume in L
                        = 0.256*0.0321 = 0.0082moles
   1 moles of NaCl react with AgNo3 to gives 1 moles of AgCl
   0.0082 moles of NaCl react with AgNO3 to gives 0.0082 moles of AgCl
   mass of AgCL = no of moles * gram molar mass
                 = 0.0082*143.5 = 1.1767g >>>>answer

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