Question

In the following reaction, how many grams of NaBr will react with 311 grams of Pb(NO3)2?

Answer 1

Molar mass :

NaBr = 103 g/mol

Pb(NO3)2 = 331.20 g/mol

Balanced chemical equation :

2 NaBr + 1 Pb(NO3)2 = 2 NaNO3 + 1 PbBr2

2*103 g NaBr ------------> 1 * 331.20 g Pb(NO3)2
g NaBr -------------------> 311 g Pb(NO3)2

331.20 g = 2*103*311

331.20 g = 64066

mass ( NaBr ) = 64066 / 331.20

mass ( naBr) = 193,43 g of NaBr

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