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 %.