The standard reduction potentials for two half-cells involving iron are: given below. Fe2+ (aq) + 2e– ® Fe (s) Eο = –0.44 V Fe3+
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I am attempting the problem for phosphonium Ion rather than its chloride salt. The chemical equation is shown below along with molar masses in mg.
First of all we will calculate the amounts of reactants required for the synthesis of 220 mg of phophonium ion. Calculations for both reactants is as follow,
For Benzyl chloride,
= 
Solving for X,
X =
X = 78.79 mg
For PPh₃:
= 
Solving for X,
X =
X = 163.27 mg
Now, Assuming these values as for 95 % conversion, we can calculate 100 % yield as follow,
when
= 
Solving for X,
X =
= 231.57 mg
Now, calculate reactants mass with respect to 231.57 mg
when
= 
Solving for ,
X =
= 82.93 mg of Benzyl chloride
when = 
Solving for ,
X =
= 171.85 mg of PPh3
So, reaction was started with reacting 82.93 mg of Benzyl Chloride and 171.85 mg of Triphenyl Phosphine.
First of all we will calculate the amounts of reactants required for the synthesis of 220 mg of phophonium ion. Calculations for both reactants is as follow,
For Benzyl chloride,
Solving for X,
X =
X = 78.79 mg
For PPh₃:
Solving for X,
X =
X = 163.27 mg
Now, Assuming these values as for 95 % conversion, we can calculate 100 % yield as follow,
when
Solving for X,
X =
Now, calculate reactants mass with respect to 231.57 mg
when
Solving for ,
X =
when
Solving for ,
X =
So, reaction was started with reacting 82.93 mg of Benzyl Chloride and 171.85 mg of Triphenyl Phosphine.