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A quality analyst mixes 3ml of 0.255M of Bacl2 with excess Agno3 causing Agcl to precipitate and subsequently isolate 2.01g of Agcl.calculate the yield of the reaction
Molar mass of BaCl2 = 208.23 g/mol0.255 × 208.23 = 53.0992/53.099 × 100= 3.7665%
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If 108.5 moles of FeCl3 is combined with 76.8 moles of AgNo3, how many moles of Agcl is produced? With solution
3AgNO3 + 3FeCl3 = 3AgCl + 3Fe(NO3)Mole ratio of FeCl3:AgCl = 1:1Mole ratio of AgNO3:AgCl = 1:1108.5+76.8 = 185.3moles
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If a 10% excess is to be used, how many milliliters of 0.15 M AgNO3 are needed to precipitate the AgCl from a sample which weighs 0.5000 g and contains 3 mM ZnCl2?
0.0033
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Suppose that 32.9 mL of 0.479 M NaCl is added to 32.9 mL of 0.331 M AgNO3.
(a) How many moles of AgCl would precipitate?
0.479 mol in 1000 mL x in 32.9 mL x=0.479*32.9/1000=0.01576 mol of NaCl 0.331 mol in 1000 ml y in 32.9 y=0.0109 mol of AgNO3 NaCl is in excess NaCl+AgNO3----gt;NaNO3...
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how many grams of AgCl will precipitate if 5.8g of 0.1M NaCl is added to the saturated soln of AgCl
Solution of AgCl is already saturated, so no extra AgCl could precipitate. I guess that saturated solution of AgNO3 was meant. Also AgCl can't react with NaCl, so addition of...
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will a precipitate occur if 250 ml of 1.5 x 10^.3 M NaCl is mixed with 250 mL of 2.0 x 10^.7 M AgNO3? Ksp of AgCl = 1.6 x 10^-10
The equation for the chemical reaction is NaCl(aq) + AgNO3(aq) = AgCl(s) + NaNO3(aq) Ksp = [Ag+] x [Cl-] = 1.6 x 10^-10 From the equation for Ksp: [Ag+] =...
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25.00 mL aliquot of 0.016 M NaCl was titrated with 27.52 mL of 0.18 M AgNO3. Determine whether a precipitate will form at end point. Ksp AgCl = 1.6 x10-10
Solution:Cinitial × Vinitial = Cfinal × VfinalVfinal = V(NaCl) + V(AgNO3) = 25.00 mL + 27.52 mL = 52.52 mLVfinal = 52.52 mLThe concentration of NaCl when the solutions are mixed:Cfinal(NaCl)...
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Need the ionic equation and net ionic equations
1. AgNO3(aq) +KCl(aq) ----> KNO3 + AgCl
2. CuCl2(aq) + Ba(NO3)2(aq) -----> Cu(NO3)2 + BaCl2
Thank you so much for helping me!
1. AgNO3(aq) +KCl(aq) ----> KNO3(aq) + AgCl (s) Ag+ + NO3- +K+ + Cl- ----> K+ + NO3- + AgCl(s) Ag+ + Cl- ----> AgCl(s) 2. CuCl2(aq) + Ba(NO3)2(aq) ----->...
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If 25 mL of a 0.45 M AgNO3 solution is added to 55 mL of a 1.2M NaCl solution. What mass of AgCl will be precipitated.
AgNO3 (s) + NaCl(aq) ==>> AgCl(s) + NaNO3 (aq)
Calculate moles of both: n=C×V n(AgNO_3 )=0.45 mol/L×0.025 L=0.01125 mol n(NaCl)=1.2 mol/L×0.055 L=0.066 mol NaCl is in excess, AgNO3 is limiting reagent. Thus, n(AgCl)=n(AgNO_3 )=0.01125 mol As we know: n=m/MW...
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AgNO3+NH4Cl --> AgCl+NH4NO3. If you have 50.26g of AgNO3, how many grams of NH4NO3 can be produced?
AgNO3+NH4Cl = AgCl+NH4NO3 n(AgNO3) = n(NH4NO3) n(AgNO3) = 50,26 / 169,87 = 0,296 mol = n(NH4NO3) m(NH4NO3) = n(NH4NO3) * Mr(NH4NO3) m(NH4NO3) = 0,296 * 80 = 23,67 g
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