When solutions containing 25 g each of lead(II) nitrate andpotassium iodide are

Question

When solutions containing 25 g each of lead(II) nitrate andpotassium iodide are mixed, a yellow precipitate results.
(a) what type of reaction occurred? (b) What is the name and formula for the solid product? (c) If, after the filtration and drying, the solid productweighted 7.66 g, what was the percent yield for the reaction?

(a) what type of reaction occurred? (b) What is the name and formula for the solid product? (c) If, after the filtration and drying, the solid productweighted 7.66 g, what was the percent yield for the reaction?

Explanation / Answer

The reaction taking place is given by                                               Pb( NO 3 ) 2(aq) + 2KI(aq) ==>PbI2(s) + 2KNO3(aq) (a) the reaction is a double replacement reaction. (b)name of the solid product is Lead iodide & itsformula is PbI2 (c)     Pb( NO 3 )2(aq) + 2KI(aq) ==> PbI2(s) +2KNO3(aq)      Molecular weight of Pb( NO3 ) 2 =207.2 + (2* 14 ) + ( 6* 16 )                                                           =331.2g        Molecular weight of KI = 39+126.9                                            = 165.9 g      Molecular weightof   PbI2 = 207.2 +( 2*126.9)                                               =461.0 g    Molecular weightof   KNO3 = 39+14+( 3*16)                                                =101 g   Pb( NO 3 ) 2(aq) + 2KI(aq)==> PbI2(s) + 2KNO3(aq) 331.2g                  2* 165.9 g       461 g                             ( 331.8 g ) So, 331.2 g of  Pb( NO 3 )2 upon reaction gives 461 g of PbI2           25 g of  Pb( NO 3 ) 2 upon reactiongives X g of PbI2                            X = ( 461 * 25 ) / 331.2                                = 34.7977 g So , theoritically the yield must be 34.7977 g ofPbI2 But practically the yield obtained is 7.66 g So, the percentage yield = ( (Theoritical yield -Practical yield )* 100 ) / Theoritical yield                                      = ( ( 34.7977 - 7.66 ) * 100 ) / 34.7977                                      = 77.98                            X = ( 461 * 25 ) / 331.2                                = 34.7977 g So , theoritically the yield must be 34.7977 g ofPbI2 But practically the yield obtained is 7.66 g So, the percentage yield = ( (Theoritical yield -Practical yield )* 100 ) / Theoritical yield                                      = ( ( 34.7977 - 7.66 ) * 100 ) / 34.7977                                      = 77.98

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