If 0.110 mol of liquid H2O and 40.5 g of solid Al2S3 are reacted stoichiometrically according to the balanced equation, how many moles of solid Al(OH)3 are produced?
Write the equation and balance it.
This is an "excess reagent problem."
2. Convert 40.5 g Al2S3 to moles. moles = g/molar mass.
3. Using the coefficients in the balanced equation, convert
a. moles liquid water to moles Al(OH)3.
b. moles Al2S3 to moles Al(OH)3.
c. Unless the answer to a and b are the same, one of the reagents is in excess; the smaller number of moles Al(OH)3 is the correct one to use and the OTHER reagent is the one in excess.
The reaction is:
Al2S3(s) + 6H2O(l) --> 2Al(OH)3(s) + 3H2S(g)
(40.5gAl2S3)/(150.16 g/mol) = 0.270 moles Al2S3
(0.110 mol H2O)(1 mol.Al2S3/6mol.H2O)=0.0183mol Al2S3 NEEDED. We have 0.270 moles Al2S3 AVAILABLE. So, H2O is the limiting reagent. The moles of Al(OH)3 produced is based on H2O, not Al2S3 (the excess reagent). So,
moles Al(OH)3 = (0.110 moles H2O)(2 mol.Al(OH)3 / 6 mol H2O) = ________?
To find the number of moles of solid Al(OH)3 produced, we need to first determine the stoichiometric ratio between Al2S3 and Al(OH)3 using the balanced equation.
The balanced equation for the reaction is as follows:
Al2S3 + 6H2O -> 2Al(OH)3 + 3H2S
From the balanced equation, we can see that 1 mole of Al2S3 reacts with 2 moles of Al(OH)3.
Given that we have 40.5 g of Al2S3, we need to convert this mass to moles using the molar mass of Al2S3.
The molar mass of Al2S3 is:
(2 x atomic mass of Al) + (3 x atomic mass of S)
= (2 x 26.98 g/mol) + (3 x 32.06 g/mol)
= 53.96 g/mol + 96.18 g/mol
= 150.14 g/mol
Now, we can calculate the number of moles of Al2S3:
moles of Al2S3 = mass / molar mass
moles of Al2S3 = 40.5 g / 150.14 g/mol
moles of Al2S3 ≈ 0.2697 mol
Since the reaction is stoichiometric, the stoichiometric ratio between Al2S3 and Al(OH)3 is 1:2. Therefore, the number of moles of Al(OH)3 produced is two times the number of moles of Al2S3.
moles of Al(OH)3 = 2 x 0.2697 mol
moles of Al(OH)3 ≈ 0.5394 mol
So, approximately 0.5394 moles of solid Al(OH)3 are produced.