How much potash alum is prepared by 6g Al2(SO4)3 & 1.5g K2SO4?
To calculate the amount of potash alum (KAl(SO4)2·12H2O) that can be prepared from 6g of Al2(SO4)3 and 1.5g of K2SO4, we need to determine the limiting reagent.
1. Calculate the number of moles for each of the reactants:
- Moles of Al2(SO4)3 = 6g / molar mass of Al2(SO4)3
- Moles of K2SO4 = 1.5g / molar mass of K2SO4
2. Determine the stoichiometric ratio between Al2(SO4)3 and KAl(SO4)2·12H2O:
The balanced chemical equation for the reaction is:
Al2(SO4)3 + K2SO4 + 24H2O → 2KAl(SO4)2·12H2O + 3H2SO4
From the equation, we can see that 1 mole of Al2(SO4)3 reacts with 1 mole of KAl(SO4)2·12H2O.
3. Identify the limiting reagent:
The limiting reagent is the one that produces the least amount of KAl(SO4)2·12H2O. To determine this, divide the number of moles of each reactant by their respective stoichiometric coefficients in the balanced equation.
- Moles of Al2(SO4)3 / stoichiometric coefficient of Al2(SO4)3 = x
- Moles of K2SO4 / stoichiometric coefficient of K2SO4 = y
The smaller value between x and y indicates the limiting reagent.
4. Calculate the amount of KAl(SO4)2·12H2O produced:
If Al2(SO4)3 is the limiting reagent, use its moles to calculate the amount of KAl(SO4)2·12H2O produced.
- Moles of KAl(SO4)2·12H2O = (Moles of Al2(SO4)3 / stoichiometric coefficient of Al2(SO4)3) * stoichiometric coefficient of KAl(SO4)2·12H2O
- Amount of KAl(SO4)2·12H2O = Moles of KAl(SO4)2·12H2O * molar mass of KAl(SO4)2·12H2O
Similarly, if K2SO4 is the limiting reagent, use its moles to calculate the amount of KAl(SO4)2·12H2O produced.
Note: Additional steps may be needed to deal with the reaction's water of hydration, which may affect the calculations if the compound's formula changes. However, for simplicity, we'll assume the hydrated form mentioned in your question.
Remember, to accurately calculate the mass of the resulting compound, you need the molar masses of Al2(SO4)3 and KAl(SO4)2·12H2O.