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Lead(ii) iodide was prepared by reacting 65.0 ml of 0.218 m pb(no3)2 with 80.0 ml of 0.265 m ki. If the actual yield of the reaction was 3.26 g, which choice is closest to the %yield of the reaction?

User Rego
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1 Answer

4 votes

The balanced chemical equation for the reaction is as follows:


Pb(NO_(3))_(2)+2KI\rightarrow PbI_(2)+2KNO_(3)

From the molarity and volume of
Pb(NO_(3))_(2) and KI, number of moles can be calculated as follows:


n=M* V

For
Pb(NO_(3))_(2) :


n=0.218 M* 65* 10^(-3)L=0.01417 mol

Similarly, for KI:


n=0.265 M* 80* 10^(-3)L=0.0212 mol

From the balanced chemical reaction, 1 mol of
Pb(NO_(3))_(2) gives 1 mole of
PbI_(2) thus, 0.01417 mol will give 0.01417 mol.

Molar mass of
PbI_(2) is 461.01 g/mol, mass can be calculated as:


m=n* M=0.01417 mol* 461.01 g/mol=6.53 g

Similarly, 2 mol of KI gives 1 mole of
PbI_(2) thus, 0.0212 mol will give
(0.0212)/(2)=0.0106 mol of
PbI_(2).

Mass can be calculated as:


m=n* M=0.0106 mol* 461.01 g/mol=4.88 g

The amount of
PbI_(2) obtained from KI is less than that from
Pb(NO_(3))_(2) thus, KI is limiting reactant and amount of
PbI_(2) obtained from KI will be theoretical yield.

Actual yield is 3.26 g, % yield can be calculated as follows:


percentage yield=(Actual yield)/(theoretical yield)* 100

Putting the values,


percentage yield=(3.26 g)/(4.88 g)* 100=66.80%

Therefore, % yield will be 66.80%.

User Lukassz
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