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What mass of precipitate (in g) is formed when 58.2 mL of 0.500 M FeCl₃ reacts with excess AgNO₃ in the following chemical reaction? FeCl₃(aq) + 3 AgNO₃(aq) → 3 AgCl(s) + Fe(NO₃)₃(aq)
Mass of precipitate =2.37g
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When 15.25 grams of A reacts with 34.22 grams of B, 18.69 grams of C is produced. How many grams of D are produced?
30.78 g
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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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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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A 60g sample of iron reacts with 26g
of oxygen to form how many grams of iron oxide?
Express your answer using two significant figures.
4Fe + 3O2 --> 2Fe2O3n=m/Mn(Fe)=60/56=1.07 moln(O2)=26/32=0.81 moln(Fe)/4 : n(O2)/3 = 0.2678 : 0.2708 --> Fe is a limiting agentn(Fe2O3)=n(Fe)/2n(Fe2O3)=1.07/2=0.535 molM(Fe2O3) = 56*2 + 16*3 = 160 g/molm(Fe2O3)= 0.535*160 = 85.6...
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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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3.38 g of AgNO3 react with BaCl2 to produce 1.43 g of AgCl according to the following balanced equation: 2AgNO3 + BaCl2 → 2 AgCl + Ba(NO3)2 What is the percent yield of AgCl in this reaction?
According to the equation, moles AgNO3 = moles AgCl.So the theoretical yield of this reaction can be calculated as: mtheor(AgCl) = (m(AgNO3) / M(AgNO3)) x M (AgCl) = (3.38g /...
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The Ksp of AgCl at 25oc is 1.0x10-10. Calculate the concentration of Ag+ and Cl- ions in a saturated solution of AgCl, and the molar solubility of AgCl ?
Conc. of Ag+ and Cl- is 10-5MSystem09:50Your answer received
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what is The Ksp of AgCl at 25 oc is 1.0x10-10. Calculate the concentration of Ag+ and Cl- ions in a saturated solution of AgCl, and the molar solubility of AgCl ?
AgCl <=> Ag+ + Cl-[Ag+] = [Cl-]Ksp = [Ag+][Cl-] = [Ag+]2 = 1.0×10-10[Ag+] = 1.0×10-5 M[Ag+] = [Cl-] = 1.0×10-5 MMolar solubility = [Ag+] = 1.0×10-5 M
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50 ml of HCl solution (1 mol/L) were added to 53 g of AgNO3 solution (0.22 w/w%). How many grams of AgCl were produced? The molecular masses are 169.9 g/mol (AgNO3) and 143.3 g/mol (AgCl).
HCl + AgNO3----------- AgCl + HNO3To get our limiting reagent that will be used for the reaction:Using n=CV for HclC=1mol/L V=50/1000= 0.05Ln=1×0.05=0.05molesTherefore using a mole-mass relationship If 1mole of HCl...
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