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Question

Pb(NO3)2 and Kl reacts in aqueous solution to form an yellow precipitate of PbI2. In one series of experiments, the masses of two reactants varied but the total mass of the two was held constant at 5.0 g. What maximum mass in grams of PbI2 can be produced in the above experiment? (Write your answer to the nearest integer.)

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Solution

Pb(NO3)2+2KIPbI2+2KNO3
The molar masses of Pb(NO3)2 and KI are 331.2 g/mol and 166 g/mol respectively. The mass of PbI2 is 461 g/mol.
Let X moles of Pb(NO3)2 react with 2X moles of KI to produce X moles of PbI2.
Total mass of Pb(NO3)2 and KI that has reacted is 331.2X+2×166×X=663.2X g. The total mass of two was held constant at 5.0 g

633.2X=5
X=0.007539 moles.
This is the number of moles of PbI2 obtained. The mass of PbI2 obtained =0.007539mol×461g/mol=3.4g.
The answer to the nearest integer is 3.

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