How much hydrogen chloride is required for the reaction with 150 g of aniline?

1. To solve, we write down the process:
m = 150 g. X l. -?
C6H5NH2 + HCl = C6H5NH3Cl – phenylamine chloride obtained;
1 mol 1 mol.
2. Let’s make the calculations:
M (C6H5NH2) = 93 g / mol;
Y (C6H5NH2) = m / M = 150/93 = 1.6 mol;
Y (HCl) = 1.6 mol, since the amount of substances according to the equation is 1 mol.
3. Find the volume of the original substance:
V (HCl) = 1.6 * 22.4 = 35.84 liters.
Answer: you need hydrogen chloride with a volume of 35.84 liters.