How many grams of chlorine can be liberated from the decomposition of 64.0g of AuCl3 by the following reaction: 2 AuCl3 > 2 Au + 3Cl2?
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This seems like a very costly way to produce chlorine. I've seen people discard the mixture down the sink after they're done.
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To find out how many grams of chlorine can be liberated from the decomposition of 64.0g of AuCl3, we need to use stoichiometry.

Calculate the molar mass of AuCl3: Au: 197.0 g/mol Cl: 35.45 g/mol x 3 = 106.35 g/mol Total molar mass = 197.0 g/mol + 106.35 g/mol = 303.35 g/mol

Determine the number of moles of AuCl3: Number of moles = Mass / Molar mass Number of moles = 64.0 g / 303.35 g/mol ≈ 0.211 moles

Use the stoichiometric coefficients from the balanced equation to find the moles of Cl2 produced: From the balanced equation, 2 moles of AuCl3 produce 3 moles of Cl2. So, 0.211 moles of AuCl3 will produce: (0.211 moles × 3 moles Cl2) / 2 moles AuCl3 = 0.3165 moles of Cl2

Convert moles of Cl2 to grams: Mass = Number of moles × Molar mass of Cl2 Mass = 0.3165 moles × 70.90 g/mol (molar mass of Cl2) Mass ≈ 22.4 grams
Therefore, approximately 22.4 grams of chlorine can be liberated from the decomposition of 64.0 grams of AuCl3.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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