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4 months 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:

castortr0y [3K]4 months ago

5 0

Answer:

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

Explanation:

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

To find the molar mass, we have: MW GaBr₃ = 309.4 g/mol and MW AgBr = 187.8 g/mol.

Using these values, we note that 187.8 g of AgBr corresponds to 1 mol.

If 0.368 g of AgBr yields x moles, we calculate x to be 2.0 x 10⁻³ mol AgBr.

Based on the stoichiometry, 1 mol of GaBr₃ produces 3 mol of AgBr, giving us:

y ___________ 2.0 x 10⁻³ mol AgBr

Thus, y equals 6.7 x 10⁻⁴ mol GaBr₃.

Additionally, 1 mol of GaBr₃ has a mass of 309.4 g.

Consequently, 6.7 x 10⁻⁴ mol of GaBr₃ corresponds to w grams of GaBr₃, yielding w = 0.206 g GaBr₃.

Relating this to the total mass:

0.6813 g _____ 100%

0.206 g _____ z

Therefore, z is calculated as 30.2 %.

KiRa [2.9K]4 months ago

3 0

Answer:

29.6%

Explanation:

We start with the balanced equation as follows:

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

From this equation, we can derive the following important values:

The molar mass of AgBr is given as 187.77 g/mol.
The molar ratio between AgBr and GaBr₃ is 3:1.
The molar mass of GaBr₃ registers at 309.44 g/mol.

The computation for the mass of GaBr₃ corresponding with 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 calculation for the mass percentage of GaBr₃ in the total mass of 0.6813 g is given by:

(0.202g/0.6813g) × 100% = 29.6%

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