How much potash alum is prepared by 6g Al2(SO4)3 & 1.5g K2SO4?
To determine the amount of potash alum (KAl(SO4)2) that can be prepared from 6g of Al2(SO4)3 and 1.5g of K2SO4, we need to find the limiting reactant.
1. Start by finding the molar masses of each compound:
- Al2(SO4)3: 2(26.98 g/mol) + 3(32.07 g/mol) + 12(16.00 g/mol) = 342.15 g/mol
- K2SO4: 2(39.10 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 174.26 g/mol
2. Use the molar masses to calculate the number of moles for each compound:
- Moles of Al2(SO4)3 = Mass / Molar mass = 6g / 342.15 g/mol
- Moles of K2SO4 = Mass / Molar mass = 1.5g / 174.26 g/mol
3. Determine the molar ratio between Al2(SO4)3 and KAl(SO4)2, which is 1:1. This means that for every 1 mole of Al2(SO4)3, we need 1 mole of KAl(SO4)2.
4. Compare the moles of Al2(SO4)3 and K2SO4 to determine the limiting reactant:
- Divide the moles of Al2(SO4)3 by the molar ratio (1:1) to get moles of KAl(SO4)2 that can be formed.
- Divide the moles of K2SO4 by the molar ratio (1:1) to get moles of KAl(SO4)2 that can be formed.
The smaller value obtained in step 4 represents the limiting reactant. It determines the maximum amount of product (potash alum) that can be formed.
5. Once we've identified the limiting reactant, we can use its moles to calculate the amount (in grams) of potash alum formed.
- Multiply the moles of KAl(SO4)2 by its molar mass to determine the mass of potash alum formed.
Remember, the theoretical yield assumes 100% efficiency. In practice, the actual yield may be less due to factors such as incomplete reactions or side reactions.
So, now you have the step-by-step explanation to calculate the amount of potash alum that can be prepared from 6g of Al2(SO4)3 and 1.5g of K2SO4.