Question

According to the following reaction, what amount of Al2S3 remains when 20.00 g of Al2S3 and 2.00 g of H2O are reacted? A few of the molar masses are as follows: Al2S3 = 150.17 g/mol, H2O = 18.02 g/mol.

Al2S3(s) + 6 H2O(l) ? 2 Al(OH)3(s) + 3 H2S(g)

a)14.00 g
b)8.33 g
c)19.78 g
d)17.22 g
e)28.33 g

Answer 1

First you need your mole ratio,
2g of H2O / (1mol/18g H2O) = 0.11 moles

20g Al2S3 / (1mol/150.17g) = 0.13 moles

Therefore H2O is your limiting reagent.

(2.0g H2O X 150.17g Al2S3)/(18 X6) = mass of Al2S3= 2.78g of Al2S3


That is the amount of Al2S3 you used, then 20-2.78=17.22g

Answer 2

d)17.22 g

Answer 3

The answer is: d)17.22 g Al253 (s) 6 H20()>2 Al(OH)3(s) 3 H25(g) Moles of H2O = mass/molar mass of H2O = 2.00/18.02 0.11 10 mol Moles of A1253 reacted 1 /6 x moles of H2O = 1 /6 x 0.1110 = 0.01 850 mol Mass of AI2S3 reacted moles x molar mass of AI253 0.01850 x 150.17 2.78 g Mass of Al253 remaining 20.00 2.78 17.22 g