For each product of the following unbalanced equations which reactant is limiting, and calculate the mass of each product that is expected.
1) 2Al + 6HCL --> 2AlCl2 + 3H2
15.0gAl * 1moleAl/26.981gAl * 2moleAlCl3/2moleAl * 133.34gAlCl3/1moleAlCl3= 74.1gAlCl3
15.0gHCL * 1moleHCL/36.461gHCL * 2moleAlCl3/6moleHCL * 133.34gAlCl3/1moleAlCl3 = 18.3gAlCl3
The second equation would be the Limiting reaction. I have no clue how to solve the rest of this problem.
You seem to have mixed up AlCl2 and the real product, AlCl3. Your equation would be balanced if you had written the aluminum chloride formula correctly.
To find out which is the limiting reactant, compute the number of moles available of each reactant. You have 0.556 moles of Al and 0.441 moles of HCl. Since there are not three times (or more) as many moles of HCl, HCl is the limiting reactant.
To determine the limiting reactant and calculate the mass of each product, you need to compare the number of moles of each reactant to the stoichiometric ratio given in the balanced equation.
1) First, calculate the number of moles for each reactant:
Number of moles of Al = mass of Al / molar mass of Al
= 15.0 g Al / 26.981 g/mol Al
= 0.556 mol Al
Number of moles of HCl = mass of HCl / molar mass of HCl
= 15.0 g HCl / 36.461 g/mol HCl
= 0.411 mol HCl
2) Next, use the stoichiometric ratio from the balanced equation to determine how many moles of each product can be formed:
From the balanced equation: 2 Al + 6 HCl -> 2 AlCl2 + 3 H2
According to the stoichiometry:
- For every 2 moles of Al, we expect 2 moles of AlCl2 to be produced.
- For every 6 moles of HCl, we expect 2 moles of AlCl2 to be produced.
3) Now, determine which reactant is the limiting reactant:
- The stoichiometry tells us that for every 2 moles of Al, we expect to produce 2 moles of AlCl2.
- However, we only have 0.556 moles of Al, which is less than the stoichiometric amount needed (2 moles for each 2 moles of AlCl2).
- On the other hand, we have 0.411 moles of HCl, which is more than the stoichiometric amount needed (6 moles for each 2 moles of AlCl2).
Since the amount of HCl is limited by the stoichiometry, HCl is the limiting reactant.
4) Finally, calculate the mass of each product that is expected:
AlCl2:
Number of moles of AlCl2 = limited reactant (HCl) moles * stoichiometric ratio of AlCl2 / HCl
= 0.411 mol HCl * (2 mol AlCl2 / 6 mol HCl)
= 0.137 mol AlCl2
Mass of AlCl2 = number of moles of AlCl2 * molar mass of AlCl2
= 0.137 mol AlCl2 * 133.34 g/mol AlCl2
= 18.3 g AlCl2
H2:
Number of moles of H2 = limited reactant (HCl) moles * stoichiometric ratio of H2 / HCl
= 0.411 mol HCl * (3 mol H2 / 6 mol HCl)
= 0.205 mol H2
Mass of H2 = number of moles of H2 * molar mass of H2
= 0.205 mol H2 * 2.016 g/mol H2
= 0.413 g H2
So, the expected mass of AlCl2 is 18.3 g and the expected mass of H2 is 0.413 g.