Calculate the mass of silver bromide produced from 22.5 g of silver nitrate.

Equation is as follow,

2 AgNO ₃ + MgBr ₂ → 2 AgBr + Mg (NO ₃) ₂

According to eq.

339.74 g (2 moles) AgNO₃ produces = 375.54 g (2 moles) of AgBr

So,

22.5 g AgNO₃ will produce = X g of AgBr

Solving for X,

X = (22.5 g * 375.54 g) : 339.74 g

X = 24.87 g of AgBr