Find the mass of salt formed by passing excess hydrogen bromide through a solution containing 80.1 grams of alanine.

First, we write down the equation of reactions.
CH3CH (NH2) COOH + HBr = [CH3CH (NH3) COOH] Br.
Next, we calculate the molar mass of alanine.
M (alanine) = 12 × 3 + 7 + 14 + 16 × 2 = 89 g / mol alanine.
Next, we find the molar mass of the salt obtained.
M (salt) = 79.9 + 1 + 89 = 169.9 g / mol of salt.
Further, according to the reaction equation, we find the mass of the obtained salt.
80.1 g alanine – x g salt
89 g / mol alanine – 169.9 g / mol salt
X = 80.1 × 169.9 ÷ 89 = 152.91 g of salt.
This means that the mass of the formed salt is 152.91 g.
Answer: m (salt) = 152.91 g.