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

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