Question

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

Answer 1

Equation is as follow,

                     2 AgNO₃  +  MgBr₂    →    2 AgBr  +  Mg(NO₃)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