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

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 %.

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:

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%