Question

Consider the following precipitation reaction: 2 NaBr(aq) + Pb(NO3)2 (aq) → PbBr2 (s) + 2 KNO3...

Consider the following precipitation reaction: 2 NaBr(aq) + Pb(NO3)2 (aq) → PbBr2 (s) + 2 KNO3 (aq) Calculate the mass of NaBr(aq) in grams required to react with 58.1 g of Pb(NO3)2 (aq)

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Answer #1

Consider the reaction

Here 2 mole of NaBr is required to react with 1 mole Pb(NO3)2 to give the products.

As we know 1 mole is equal to the gram molar mass.

Here 2 mole of NaBr that means

and 1 mole of   Pb(NO3)2 that means

Here 102.9gm and 331.2gm are the gm molar mass of NaBr and Pb(NO3)2 respectively.

As per the question we can say:

1 mole or of 331.2gm of Pb(NO3)2 needs2 mole NaBr that means 205.8 gm of NaBr to give the products.

Then 58.1gm of Pb(NO3)2 will need

Hence the required mass of NaBr is 36.101g.

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