Based on the balanced equation

Al2S3 + 6H2O → 2Al(OH)3 + 3H2S
calculate the number of excess reagent units remaining when 4 Al2S3 formula units and 30 H2O molecules react?

Molar Mass (g/mol)
Al2S3 150.16
H2O 18.015
Al(OH)3 75.361
H2S 34.082
Avogadro's No.
6.022×1023 mol-1

To determine the excess reagent units remaining, we first need to calculate the moles of each reactant.

1. Determine the moles of Al2S3:
Moles of Al2S3 = (4 mol Al2S3) x (1 mol Al2S3/150.16 g Al2S3)
Moles of Al2S3 = 0.02664 mol Al2S3

2. Determine the moles of H2O:
Moles of H2O = (30 mol H2O) x (1 mol H2O/18.015 g H2O)
Moles of H2O = 1.6648 mol H2O

Now we need to determine the limiting reactant.

3. Calculate the mole ratio of Al2S3 to H2O based on the balanced equation:
Mole ratio of Al2S3 to H2O = (1 mol Al2S3/6 mol H2O)

4. Compare the actual mole ratio of Al2S3 to H2O:
Actual mole ratio of Al2S3 to H2O = (0.02664 mol Al2S3) / (1.6648 mol H2O)
Actual mole ratio of Al2S3 to H2O = 0.016 mol Al2S3 / mol H2O

The actual mole ratio of Al2S3 to H2O is less than the mole ratio based on the equation, which means Al2S3 is the limiting reactant.

5. Calculate the moles of Al(OH)3 formed:
Moles of Al(OH)3 formed = (0.02664 mol Al2S3) x (2 mol Al(OH)3 / 1 mol Al2S3)
Moles of Al(OH)3 formed = 0.05328 mol Al(OH)3

6. Calculate the moles of H2S formed:
Moles of H2S formed = (0.02664 mol Al2S3) x (3 mol H2S / 1 mol Al2S3)
Moles of H2S formed = 0.07992 mol H2S

7. Determine the excess reagent units remaining:
Excess reagent units remaining = Moles of H2O - Moles of Al(OH)3 formed
Excess reagent units remaining = 1.6648 mol H2O - 0.05328 mol Al(OH)3
Excess reagent units remaining = 1.61152 mol H2O

Now, to convert the excess reagent units remaining into molecules of H2O:
(Number of molecules in 1 mol H2O) x (Excess reagent units remaining) x (Avogadro's number)
= (6.022 x 10^23 molecules/mol) x (1.61152 mol H2O)

Excess reagent units remaining = 9.7044 x 10^23 molecules H2O

Therefore, when 4 Al2S3 formula units and 30 H2O molecules react, there are approximately 9.7044 x 10^23 excess reagent units of H2O remaining.

To calculate the number of excess reagent units remaining, we need to determine which reactant is the limiting reagent and the stoichiometry of the reaction.

First, we determine the moles of each reactant using the given information:

For Al2S3:
Molar mass of Al2S3 = 150.16 g/mol
Number of Al2S3 formula units given = 4
Moles of Al2S3 = (4 formula units) / (6.022×10^23 formula units/mol) = 6.645×10^-24 mol

For H2O:
Molar mass of H2O = 18.015 g/mol
Number of H2O molecules given = 30
Moles of H2O = (30 molecules) / (6.022×10^23 molecules/mol) = 4.979×10^-23 mol

Next, we compare the mole ratio of Al2S3 to H2O in the balanced equation to determine the limiting reagent:

From the balanced equation, the mole ratio of Al2S3 to H2O is 1:6.

Moles of H2O / Moles of Al2S3 = (4.979×10^-23 mol) / (6.645×10^-24 mol) = 7.5
Since this ratio is greater than 6, H2O is in excess.

Therefore, Al2S3 is the limiting reagent.

To determine the number of excess reagent units remaining, we calculate how many moles of Al(OH)3 and H2S are formed using the balanced equation:

From the balanced equation, the mole ratio of Al2S3 to Al(OH)3 is 2:2, and the mole ratio of Al2S3 to H2S is 1:3.

Moles of Al(OH)3 formed = 2 * Moles of Al2S3 = 2 * 6.645×10^-24 mol = 1.329×10^-23 mol

Moles of H2S formed = 3 * Moles of Al2S3 = 3 * 6.645×10^-24 mol = 1.994×10^-23 mol

Finally, we calculate the moles of the excess reagent remaining, which is H2O:

Moles of H2O remaining = Moles of H2O initially - Moles of H2O used = 4.979×10^-23 mol - 6 * (6.645×10^-24 mol) = 4.625×10^-23 mol

To convert moles of H2O remaining to units (molecules), we use Avogadro's number:

Number of H2O molecules remaining = Moles of H2O remaining * (6.022×10^23 molecules/mol) = 4.625×10^-23 mol * (6.022×10^23 molecules/mol) = 2.7849 * 10^9 molecules

Therefore, when 4 Al2S3 formula units and 30 H2O molecules react, there are approximately 2.7849 * 10^9 H2O molecules remaining as excess reagent units.

This is a limiting reagent problem. Step 1 is to determine the limiting reagent; i.e., is it Al2S3 or H2O? You determine this from looking at the coefficients in the balanced equation. Just pick one. I'll pick H2O first, then do Al2S3.

The coefficients tell us that 6 units of H2O require 1 unit of Al2S3; therefore, 30 units of water (given in the problem) will require 7.5 units of Al2S3 and the way you get that is
30 units H2O x (1 unit Al2S3/4 units H2O) = 7.5 units Al2S3. Note that I arranged the factor (1 unit Al2S3/4 units H2O) so that the factor converts units of H2O to units of Al2S3 by making the units H2O in the first term cancel with units H2O in the denominator of the factor). So, do we have 7.5 units of Al2S3. No, the problem says we have 4 so we know Al2S3 is the limiting reagent BUT let's do it the same way and check it.
4 units Al2S3 x (6 units H2O/1 unit Al2S3) = 24 units H2O. Do we have 24? Yes, the problem says we have 30 so Al2S3 indeed is the limiting reagent, it will require 24 units of H2O to use all 4 units of Al2S3, which means 30-24=6 units of water will remain after the reaction is complete. Check my work.