What is the mass of silver in 3.4 g AgNo3?

The answer is 2.2 g.

First, we need to calculate a molar mass (M) of AgNO₃. The molar mass is a mass of 1 mole of a substance. It is the sum of relative atomic masses (Ar), which are masses of atoms of the elements.

Relative atomic mass of Ag: Ar (Ag) = 108 g

Relative atomic mass of N: Ar (N) = 14 g

Relative atomic mass of O: Ar (O) = 16 g

So, the molar mass of AgNO₃ is:

M (AgNO₃) = Ar (Ag) + Ar (N) + 3Ar (O) = 108 + 14 + 3*16 = 170 g

Now, we can make a proportion:

If 108 g of Ag is in 170 g of AgNO₃, how many silver will be in 3.4 g of AgNO₃:

108 : 170 = x : 3.4

After crossing the products:

x = 108 * 3.4 : 170 = 2.16 g ≈ 2.2 g