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

pb (no3) 2 (aq) + 2NaBr (g) - > PbBr2 (s) + 2NaNO3 (aq)

The molar mass of NaBr is 102.9 grams and that of Pb (NO3) 2 is 331.21 grams.

Pb (NO3) 2 (aq) + 2NaBr (g) → PbBr2 (s) + 2NaNO3 (aq)

mol Pb (NO3) 2 available = 311 g

Pb (NO3) 2 x [1 mol / 331.21 g] = 0.939 mol Pb (NO3) 2 (according to equation)

1 mole of Pb (NO3) 2 requires 2 moles of NaBr so:

0.939 mol Pb (NO3) 2 reacts with 0.939 x 2 = 1.88 mol NaBr

Therefore:

Mass of NaBr required = 1.88 mol NaBr x (102.9 g / mol) = 193 g NaBr