To calculate the number of grams of Al₂S₃ that can be prepared, we first need to determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
Let's start by calculating the number of moles for each reactant using their respective molar masses:
Molar mass of Al: 26.98 g/mol Molar mass of S: 32.07 g/mol
Number of moles of Al = mass of Al / molar mass of Al Number of moles of Al = 20 g / 26.98 g/mol ≈ 0.741 mol
Number of moles of S = mass of S / molar mass of S Number of moles of S = 30 g / 32.07 g/mol ≈ 0.935 mol
Next, we'll determine the stoichiometric ratio of Al₂S₃ to Al and S from the balanced chemical equation. The balanced equation for the reaction is:
2 Al + 3 S → Al₂S₃
According to the balanced equation, 2 moles of Al react with 3 moles of S to produce 1 mole of Al₂S₃.
Now, we'll determine the limiting reactant:
The ratio of Al to S in the balanced equation is 2:3. Therefore, for every 2 moles of Al, we need 3/2 moles of S. Let's calculate how many moles of S would be required to react with 0.741 moles of Al:
Moles of S required = (3/2) * 0.741 mol = 1.111 mol
Since we only have 0.935 moles of S available, it is the limiting reactant. Therefore, it will determine the amount of Al₂S₃ that can be formed.
To calculate the number of moles of Al₂S₃ formed, we use the stoichiometric ratio from the balanced equation:
Number of moles of Al₂S₃ = 0.935 mol (since S is the limiting reactant)
Finally, we'll calculate the mass of Al₂S₃ formed using its molar mass:
Molar mass of Al₂S₃: 150.16 g/mol
Mass of Al₂S₃ = Number of moles of Al₂S₃ * molar mass of Al₂S₃ Mass of Al₂S₃ = 0.935 mol * 150.16 g/mol ≈ 139.98 g
Therefore, the number of grams of Al₂S₃ that can be prepared is approximately 139.98 g.
To determine the amount of excess reactant, we can compare the moles of the non-limiting reactant (Al) with the stoichiometric ratio.
The stoichiometric ratio from the balanced equation shows that 2 moles of Al are required to react with 3 moles of S to produce 1 mole of Al₂S₃.
Since we have 0.741 moles of Al, we can calculate the moles of S needed for this amount of Al:
Moles of S needed = (3/2) * 0.741 mol = 1.111 mol
However, we only have 0.935 mol of S available. The difference between the required moles and the available moles of S is:
Excess moles of S = Moles of S available - Moles of S needed Excess moles of S = 0.935 mol - 1.111 mol ≈ -0.176 mol
The negative value indicates that there is a deficiency of S, not an excess. Therefore, there is no excess reactant in this case.