How many grams of chlorine can be liberated from the decomposition of 52.0 g of AuCl3 by this reaction? 2 AuCl3 ---> 2 Au + 3 Cl2?

Molar mass of AuCl₃ = (197.0 + 35.5×3) g/mol = 303.5 g/mol
Molar mass of Cl₂ = 35.5 × 2 = 71.0 g/mol

2 AuCl₃ → 2 Au + 3 Cl₂
Mole ratio AuCl₃ : Cl₂ = 2 : 3

No. of moles of AuCl₃ decomposed = (52.0 g) / (303.5 g/mol) = 0.171 mol
No. of moles of Cl₂ liberated = (0.171 mol) × (3/2) = 0.257 mol
Mass of Cl₂ liberated = (0.257 mol) × (71.0 g/mol) = 18.2 g

Molar mass of AuCl3 = 303.3256 g/mol
Mol AuCl3 in 52.0g = 52.0/303.3256 = 0.1714 mol
From the equation
2mol AlCl3 produce 3 mol Cl2
0.1714 mol AlCl3 will produce 0.1714*3/2 = 0.2571 mol Cl2
Molar mass Cl2 = 35.45*2 = 70.9g/mol
Mass of 0.2571 mol Cl2 = 0.2571*70.9 = 18.2g Cl2 produced ( 3 significant digits)

...

. . . . 52.0g . . . . . . . . . . . . . . . . . 18.2g [[ANSWER]]
. . . 2AuCl3 . . → . . 2Au . . + . . 3Cl2
(303.325g/mol) . . . . . . . . . . . (70.906g/mol)
. 0.1714 mol . . . . . . . . . . . . . 0.25715

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2AuCl₃ ---> 2Au + 3Cl₂

atomic mass
Au = 197
Cl = 35.5
2AuCl₃ = 2•(197+3• 35.5) = 607
2Au = 394
3Cl₂ = 213

check 607 = 394 + 213

607 grams of AuCl₃ ➜ 394 grams of Au + 213 grams of Cl₂
2 moles of AuCl₃ ➜ 2 moles of Au + 3 moles of Cl₂

"How many grams of chlorine can be liberated from the decomposition of 52.0 g of AuCl₃"

the ratio is 213/607 = x/52
x = 52• 213/607 = 18.2 g

note that I used rounded values for atomic mass. You can get a more accurate
value if you use more precision.