how many grams of chlorine can be liberated from the decomposition of 64.0g of AuCl3 by this reaction
2 AuCl3----> 2 Au + 3cl2
To determine the number of grams of chlorine liberated from the decomposition of 64.0g of AuCl3, we need to use stoichiometry and the molar masses of the compounds involved.
First, let's calculate the molar mass of AuCl3:
Atomic mass of Au = 196.97 g/mol
Atomic mass of Cl = 35.45 g/mol
Molar mass of AuCl3 = 196.97 + (3 * 35.45) = 303.32 g/mol
Next, we need to calculate the moles of AuCl3 in 64.0g:
Moles of AuCl3 = mass / molar mass = 64.0g / 303.32 g/mol
Now, we can use the balanced equation to determine the molar ratio between AuCl3 and Cl2:
2 AuCl3 -> 2 Au + 3 Cl2
From the balanced equation, we can see that for every 2 moles of AuCl3, we get 3 moles of Cl2.
Therefore, we can calculate the moles of Cl2 liberated:
Moles of Cl2 = (moles of AuCl3) * (3 moles of Cl2 / 2 moles of AuCl3)
Finally, we can calculate the grams of Cl2 liberated using the molar mass of Cl2 (70.91 g/mol):
Grams of Cl2 = moles of Cl2 * molar mass of Cl2
By plugging in the values, you can calculate the final result.
To determine the number of grams of chlorine that can be liberated from the decomposition of 64.0g of AuCl3, we need to use stoichiometry.
1. Start by writing down the balanced equation for the reaction:
2 AuCl3 -> 2 Au + 3 Cl2
2. Convert the given mass of AuCl3 to moles. To do this, divide the given mass by the molar mass of AuCl3. The molar mass of AuCl3 can be calculated by adding the atomic masses of gold (Au) and chlorine (Cl).
Molar mass of AuCl3 = atomic mass of Au + 3 * atomic mass of Cl
= (197.0 g/mol) + 3 * (35.45 g/mol)
= 303.35 g/mol
64.0 g AuCl3 * (1 mol AuCl3 / 303.35 g AuCl3) = 0.211 mol AuCl3
3. Now we need to use the stoichiometric coefficients from the balanced equation to determine the moles of Cl2 produced from the given amount of AuCl3. From the balanced equation, we can see that 3 moles of Cl2 are produced for every 2 moles of AuCl3.
Therefore, the moles of Cl2 = (0.211 mol AuCl3) * (3 mol Cl2 / 2 mol AuCl3) = 0.3165 mol Cl2
4. Finally, we can convert the moles of Cl2 to grams by multiplying by the molar mass of Cl2.
Molar mass of Cl2 = 2 * atomic mass of Cl
= 2 * 35.45 g/mol
= 70.90 g/mol
0.3165 mol Cl2 * (70.90 g Cl2 / 1 mol Cl2) = 22.48 g Cl2
Therefore, approximately 22.48 grams of chlorine can be liberated from the decomposition of 64.0 grams of AuCl3 by the given reaction.
wrong
All you need to do is convert to moles AuCL3 and then use the mole ratio and convert back to grams Cl2
64.0*(1 mol/233g)*(3 mol Cl2/2mol AuCl3)*70.906gCl2=
29.214