Respuesta :

transitionstate
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
Oseni

The mass of silver bromide produced from 22.5 g silver nitrate would be 50.01 grams

Stoichiometric calculation

From the balanced equation of the reaction:

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

The mole ratio of AgNO3 and AgBr is 2:1

Mole of 22.5 g AgNO3 = 22.5/169.87

                                       = 0.1325 mole

Equivalent mole of AgBr = 0.265 mole

Mass of 0.265 mole AgBr = 0.265 x 188.77

                                           = 50.01 grams

More on stoichiometric calculation can be found here: https://brainly.com/question/8062886