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.
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?
2 answers
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) = ________?
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) = ________?
2 answers