To a solution containing 16.2 grams of hydrogen bromide was added 6 grams of sodium hydroxide.

To a solution containing 16.2 grams of hydrogen bromide was added 6 grams of sodium hydroxide. Calculate the mass of sodium bromide that can be isolated and the resulting solution

To solve, we compose the reaction equation:
HBr + NaOH = NaBr + H2O – ion exchange reaction, sodium bromide salt was obtained;
M (HBr) = 80.9 g / mol;
M (NaOH) = 39.9 g / mol;
M (NaBr) = 102.8 g / mol;
Determine the amount of moles of hydrogen bromide, sodium hydroxide:
Y (HBr) = m / M = 16.2 / 80.9 = 0.237 mol;
Y (NaOH) = m / M = 6 / 39.9 = 0.15 mol (deficient substance);
Calculations are made for the substance in deficiency:
0.15 mol (NaOH) – X mol (NaBr);
-1 mol – 1 mol from here, X mol (NaBr) = 0.15 * 1/1 = 0.15 mol;
Find the mass of sodium bromide:
m (NaBr) = Y * M = 0.15 * 102.8 = 15.42 g.
Answer: the mass of sodium bromide salt is 15.42 g.