Complete a reaction table (in moles) for the reaction of 2.17 mol Al and 4.97 mol HCl.
2Al + 6HCl 2AlCl3 + 3H2
initial 2.17 4.97 0 0
delta ? -4.97 ? ?
final ? 0 ? ?
How do you get the question marks? Please help and explain if you can!
2Al + 6HCl 2AlCl3 + 3H2
initial 2.17 4.97 0 0
delta ? -4.97 ? ?
final ? 0 ? ?
How do you get the question marks? Please help and explain if you can!
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What's happening is that an equilibrium is forming. So from the start, having placed these compounds in a container, you're going to subtract _x moles from them, _ being their coefficients, and add _x to the other side.
2Al + 6HCl ---------> 2AlCl3 + 3H2
-2x -6x +2x +3x
You can find x because you have 6x already. 6x = 4.97, so x = 0.828
-2(0.828) -4.97 +2(0.828) +3(0.828)
Subtract or add these numbers from reactants and products, and at equilibrium you'll have the following
0.51 0 1.66 2.48
2Al + 6HCl ---------> 2AlCl3 + 3H2
-2x -6x +2x +3x
You can find x because you have 6x already. 6x = 4.97, so x = 0.828
-2(0.828) -4.97 +2(0.828) +3(0.828)
Subtract or add these numbers from reactants and products, and at equilibrium you'll have the following
0.51 0 1.66 2.48