Question

A solid mixture weighs 0.6813 g. It contains gallium bromide (GaBr3) and other inert impurities. When the solid mixture was dissolved in water and treated with excess silver nitrate (AgNO3), 0.368 g of AgBr was precipitate. A balanced chemical equation describing the reaction is provided below. GaBr3(aq) + 3 AgNO3(aq) ⟶ 3 AgBr(s) + Ga(NO3)3(aq) What is the percent of mass of GaBr3 in the solid mixture?

Answer 1

The percent of mass of GaBr₃ in the solid mixture is 30.2 %.

Step-by-step explanation:

GaBr₃(aq) + 3 AgNO₃(aq) ⟶ 3 AgBr(s) + Ga(NO₃)₃(aq)

MW GaBr₃ = 309.4 g/mol

MW AgBr = 187.8 g/mol

187.8 g AgBr _______ 1 mol

0.368 g AgBr _______ x

x = 2.0 x 10⁻³ mol AgBr

1 mol GaBr₃ ____ 3 mol AgBr

y ____ 2.0 x 10⁻³ mol AgBr

y = 6.7 x 10⁻⁴ mol GaBr₃

1 mol GaBr₃ ____________ 309.4 g

6.7 x 10⁻⁴ mol GaBr₃ ______ w

w = 0.206 g GaBr₃

0.6813 g _____ 100%

0.206 g _____ z

z = 30.2 %

Answer 2

29.6%

Step-by-step explanation:

Let's consider the following balanced equation.

GaBr₃(aq) + 3 AgNO₃(aq) ⟶ 3 AgBr(s) + Ga(NO₃)₃(aq)

We can establish the following relations.

• The molar mass of AgBr is 187.77 g/mol.
• The molar ratio of AgBr to GaBr₃ is 3:1.
• The molar mass of GaBr₃ is 309.44 g/mol.

The mass of GaBr₃ that produced 0.368 g of AgBr is:

`0.368gAgBr.(1molAgBr)/(187.77gAgBr) .(1molGaBr_(3))/(3molAgBr) .(309.44gGaBr_(3))/(1molGaBr_(3)) =0.202gGaBr_(3)`

The mass percent of GaBr₃ in the 0.6813 g-sample is:

(0.202g/0.6813g) × 100% = 29.6%