how many grams of aluminum sulfate are produced if 23.33 g Al reacts with 74.44 G CuSO4
This is a limiting reagent problem. These take awhile to work but here is a worked example. Just follow the steps outlined.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
To determine the quantity of aluminum sulfate (Al2(SO4)3) produced when 23.33 g of aluminum reacts with 74.44 g of copper(II) sulfate (CuSO4), we need to follow a balanced chemical equation and use stoichiometric calculations.
The balanced equation for this reaction is:
2 Al + 3 CuSO4 -> Al2(SO4)3 + 3 Cu
Looking at the equation, we can see that for every 2 moles of aluminum (Al), 1 mole of aluminum sulfate (Al2(SO4)3) is produced.
Step 1: Convert grams of aluminum (Al) to moles using the molar mass of aluminum.
Molar mass of aluminum (Al) = 26.98 g/mol
Moles of aluminum (Al) = mass of aluminum (Al) / molar mass of aluminum (Al)
Moles of aluminum (Al) = 23.33 g / 26.98 g/mol
Moles of aluminum (Al) = 0.863 moles
Step 2: Use the stoichiometric ratio to determine the moles of aluminum sulfate (Al2(SO4)3) produced.
According to the balanced equation, 2 moles of aluminum (Al) produce 1 mole of aluminum sulfate (Al2(SO4)3).
Moles of aluminum sulfate (Al2(SO4)3) = (0.863 moles of aluminum (Al) / 2 moles of aluminum (Al)) * 1 mole of aluminum sulfate (Al2(SO4)3)
Moles of aluminum sulfate (Al2(SO4)3) = 0.4315 moles
Step 3: Convert moles of aluminum sulfate (Al2(SO4)3) to grams using the molar mass of aluminum sulfate (Al2(SO4)3).
Molar mass of aluminum sulfate (Al2(SO4)3) = 342.15 g/mol
Mass of aluminum sulfate (Al2(SO4)3) = moles of aluminum sulfate (Al2(SO4)3) * molar mass of aluminum sulfate (Al2(SO4)3)
Mass of aluminum sulfate (Al2(SO4)3) = 0.4315 moles * 342.15 g/mol
Mass of aluminum sulfate (Al2(SO4)3) = 147.90 g
Therefore, approximately 147.90 grams of aluminum sulfate (Al2(SO4)3) are produced when 23.33 grams of aluminum reacts with 74.44 grams of copper(II) sulfate (CuSO4).
To determine the number of grams of aluminum sulfate produced, we need to use the balanced chemical equation for the reaction between aluminum (Al) and copper sulfate (CuSO4):
2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu
From the balanced chemical equation, we can see that 2 moles of aluminum react with 3 moles of copper sulfate to produce 1 mole of aluminum sulfate.
To calculate the number of moles of aluminum sulfate produced, we need to follow these steps:
Step 1: Convert the given mass of aluminum (Al) to moles.
Given mass of aluminum (Al) = 23.33 g
Molar mass of aluminum (Al) = 26.98 g/mol
Number of moles of aluminum (Al) = Given mass / Molar mass = 23.33 g / 26.98 g/mol
Step 2: Determine the limiting reactant.
To determine the limiting reactant, we need to compare the number of moles of aluminum (Al) to moles of copper sulfate (CuSO4) using their respective coefficients in the balanced equation. The reactant that produces fewer moles of aluminum sulfate will be the limiting reactant.
From the balanced equation, the ratio of aluminum (Al) to copper sulfate (CuSO4) is 2:3. This means that for every 2 moles of aluminum, we need 3 moles of copper sulfate.
Number of moles of copper sulfate (CuSO4) = Number of moles of aluminum (Al) x (3 moles of CuSO4 / 2 moles of Al)
Step 3: Calculate the number of moles and grams of aluminum sulfate produced.
From the balanced equation, we know that 1 mole of aluminum sulfate (Al2(SO4)3) is produced for every 2 moles of aluminum (Al).
Number of moles of aluminum sulfate (Al2(SO4)3) = Number of moles of aluminum (Al) / 2
Finally, to determine the mass of aluminum sulfate produced:
Mass of aluminum sulfate (Al2(SO4)3) = Number of moles of aluminum sulfate x Molar mass of aluminum sulfate
Given the molar mass of aluminum sulfate (Al2(SO4)3) is necessary to complete this calculation. Please provide the molar mass of aluminum sulfate to proceed with the calculation.