Question

You have a 275 mg sample of a compound that contains aluminum and a group 7A element (F, Cl, Br, I or At). This compound reacts with excess silver nitrate as follows (“X” represents the unknown element): AlX3 + 3 AgNO3 → 3 AgX + Al(NO3)3 This reaction forms 581 mg of AgX. Identify element X.

Answer 1

molar mass of AlX3 = 27 + 3*MM(X)

mass of AlX3 = 275 mg = 0.275 g

number of mol of AlX3 = mass / molar mass = 0.275 / (27 + 3*MM(X))

molar mass of AgX = 108 + MM(X)

mass of AgX = 581 mg = 0.581 g

number of mol of AgX = mass / molar mass = 0.581 / (108 + MM(X))

AlX3 + 3 AgNO3 → 3 AgX + Al(NO3)3

from reaction above,

mol of AgX = 3*mol of AlX3

0.581 / (108 + MM(X)) = 3*0.275 / (27 + 3*MM(X))

0.581*(27 + 3*MM(X)) = 0.825*(108 + MM(X))

15.687 + 1.743*MM(X) = 89.1 + 0.825*MM(X)

0.918*MM(X) = 73.413

MM(X) = 80 g/mol

This is molar mass of Br

Answer: Br