In the following reaction, how many grams of nabr will react with 311 grams of pb(no3)2?

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

hope this helps!.

Outermost (valence) electrons