calculate the number of grams of Al2S3 which can be prepared by the reaction of 20g of Al and 30gm of sulphur. how much the non_limiting reactant in excess?
To find the number of grams of Al2S3 that can be prepared, we need to determine the limiting reactant first.
1. Start by calculating the number of moles for each reactant using their respective molar masses.
Molar mass of Al (Aluminum) = 26.98 g/mol
Molar mass of S (Sulphur) = 32.06 g/mol
Number of moles of Al = mass of Al / molar mass of Al = 20 g / 26.98 g/mol
Number of moles of S = mass of S / molar mass of S = 30 g / 32.06 g/mol
2. Now, we need to calculate the stoichiometric ratio between Al and Al2S3 using the balanced chemical equation.
The balanced equation for the reaction of Al and S to form Al2S3 is:
2 Al + 3 S => Al2S3
From the balanced equation, we can see that the stoichiometric ratio between Al and Al2S3 is 2:1.
3. Determine the limiting reactant by comparing the moles ratio of the reactants to the stoichiometric ratio.
The reactant that has a smaller ratio compared to the stoichiometric ratio is the limiting reactant.
Moles ratio of Al to Al2S3 = (Number of moles of Al) / 2
Moles ratio of S to Al2S3 = (Number of moles of S) / 3
If the moles ratio of Al to Al2S3 is smaller than the moles ratio of S to Al2S3, then Al is the limiting reactant. Otherwise, S is the limiting reactant.
4. Once we determine the limiting reactant, we can calculate the number of moles of Al2S3 produced using the stoichiometry.
If Al is the limiting reactant:
Number of moles of Al2S3 = (Number of moles of Al) / 2
If S is the limiting reactant:
Number of moles of Al2S3 = (Number of moles of S) / 3
5. Finally, calculate the mass of Al2S3 produced using its molar mass.
Mass of Al2S3 = Number of moles of Al2S3 * molar mass of Al2S3
To determine the non-limiting reactant in excess, subtract the number of moles of the limiting reactant from the number of moles of the non-limiting reactant.
I hope this explanation helps you understand how to calculate the grams of Al2S3 and identify the non-limiting reactant in excess.