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2 years ago

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A solid mixture weighs 0.6813 g. It contains gallium bromide (GaBr3) and other inert impurities. When the solid mixture was diss

olved 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?

2 answers:

Naya [18.7K]2 years ago

5 0

Answer:

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

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 %

KATRIN_1 [288]2 years ago

3 0

Answer:

29.6%

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:

$0.368gAgBr.\frac{1molAgBr}{187.77gAgBr} .\frac{1molGaBr_{3}}{3molAgBr} .\frac{309.44gGaBr_{3}}{1molGaBr_{3}} =0.202gGaBr_{3}$

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

(0.202g/0.6813g) × 100% = 29.6%

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