Using the from problem e above, if 10.2 g of H_2S were obtained in the lab numbe
ID: 499159 • Letter: U
Question
Using the from problem e above, if 10.2 g of H_2S were obtained in the lab numbers (actual yield), what is the percent yield? _____ % How many grams of K_2C_2O_4, potassium oxalate, would contain 182 grams of pure potassium? What is the molarity of a solution made by dissolving 6.08g of sodium acetate, CH_3COONa, in water, and diluting it to a total volume of 0.750L? _____ M How many grams of strontium nitrate, Sr(NO_3)_2, are needed to prepare 3.40 times 10^3 mL of a 0.925 solution? _____ g To what volume should 50.0mL of a 6.00M HCl stock solution be diluted to obtain a 1.50M HCl solution? How much water should be used? A 38.5mL sample of 0.875M AgNO_3, solution is mixed with 51.0mL sample of 0.744M NaCl solution. The AgCl precipitated has a mass of 2.06g. Calculate the limiting reactant, the theoretical yield and percent yield. AgNO_3(aq) + NaCl_(aq) rightarrow AgCl_(s) + NaNO_3(aq)Explanation / Answer
a)
molarity of
6.08 g opf sodium acetate in V = 0.75 L
mol of NaAcetate = 82.0343 g/mol
so
mol = mass/MW = 6.08/82.0343 = 0.07411 mol of NaAc
M = mol/V = 0.07411 / 0.75 = 0.0988 M of
b)
grams of Sr(NO3)2 required fo
V = 3400 mL of M = 0.925 M solution
M = mol/V
mol = MV = (0.925)(3.400) = 3.145 mol of Sr(NO3) =required
MW = 211.63 g/mol
mass = mol*MW = 211.63 *3.145 = 665.576 gof Sr(NO3)2 required
c)
Apply Dilution law
M1¨V1 = M2V2
50*6 = 1.5*V
V = 50/1.5*6 = 200 mL
d)
V = 38.5 mL of M = 0.875 AgNO#
V = 51 mL of M = 0.744 NaCl
mol of Ag+ = MV = 38.5*0.875 = 33.6875 mmol of Ag+
mol of Cl- = MV = 51*0.744 = 37.944 mmol of Cl
clearly, Ag+ is limiting
theoretical yield --> 33.6875 mmol of Ag+ --> 33.6875 mmol of AgCl
mass = molñ*MW = 33.6875 *10^-3*143.32 =4.828092 g of AgCl are expected
% yield = real/Theoretical * 100% = 2.06/4.828092 *100 = 42.6669 %
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