How many grams of NaCl are required to precipitate most of the Ag tons from 2.50

Question

How many grams of NaCl are required to precipitate most of the Ag tons from 2.50 Times 10^2 mL of a 0.0113 M AgNO_3 solution? Write the net ionic equation for the reaction. Balance the following redox equations by the half-reaction method: H_2O_2 + Fe^3+ rightarrow Fe^3+ H_2O (in acidic solution) Cu + HNO_3 rightarrow Cu^2+ + NO + H_2O (in acidic solution) CN^- + MnO_4^ rightarrow CNO6- + MnO_2 (in basic solution) Br_2 rightarrow BrO_3^- + Br^- (in basic solution) S_2O_3^2- + I_2 rightarrow I^- + S_4O_6^2- (in acidic solution) Balance the following redox equations by the half-reaction method: Mn^2+ + H_2O_2 rightarrow MnO_2 + H_2O (in basic solution) Bi(OH)_3 + SnO_2^2- rightarrow SnO_3^2- + Bi (in basic solution) Cr_2O_7^2- + C_2O_4^2- rightarrow Cr^3+ + CO_2 (in acidic solution) ClO_3^- + Cl^ rightarrow Cl_2 + CO_2 (in acidic solution) Mn^2+ + BiO_3^- rightarrow Bi^3 + + MnO_4^- (in acidic solution)

Explanation / Answer

4.83)

m = ? NaCl

Ag ions in 2.5*10^2 ml = 250 ml

M = 0.0113 M of AgNO3

note that

AgNO3(aq) + NaCl(aq) = AgCl(s) + NaNO3(aq)

the net ionic

Ag+ + Cl- --> AgCl(s)

then

find mol of Ag+ = MV = 250*0.0113 = 2.825 mmol of Ag+

then

ratio is 1:1

so

mol of Cl- required should be 2.825 mmol ofCl-

that is

2.825 mmol of NACl = 2.825*10^-3 mol of NaCl

mass = mol*MW = (2.825*10^-3)(58.0) =0.16385 g of NaCl

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