How much Al is needed to react with 675
mL of 0.655M HCl solution according to the
reaction => 2Al + 6HCl→2AlCl3+ 3H2
Help your friend!
moles HCL=.675*.655
so you need 1/3 of that in Al
change that to grams.
thank you!
To determine the amount of Al needed to react with 675 mL of 0.655 M HCl solution, we can use the stoichiometry of the balanced chemical equation and the concept of molarity.
First, let's calculate the moles of HCl present in the solution:
Moles of HCl = volume (in liters) × concentration
Moles of HCl = 0.675 L × 0.655 mol/L
Moles of HCl = 0.441375 mol
Using the balanced chemical equation:
2Al + 6HCl → 2AlCl3 + 3H2
We can see that 2 moles of Al react with 6 moles of HCl. This means that the moles of Al required should be in a 1:3 ratio with the moles of HCl.
Since we have 0.441375 mol of HCl (from the previous step), we need to find the corresponding moles of Al:
Moles of Al = (moles of HCl × 2) / 6
Moles of Al = (0.441375 mol × 2) / 6
Moles of Al = 0.147125 mol
Now, we have calculated the moles of Al needed to react with the HCl solution. To find the mass of Al required, we need to use the molar mass of Al, which is approximately 26.98 g/mol.
Mass of Al = moles of Al × molar mass of Al
Mass of Al = 0.147125 mol × 26.98 g/mol
Mass of Al = 3.96 g
Therefore, approximately 3.96 grams of Al are needed to react with 675 mL of a 0.655 M HCl solution.