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.

Mass of Al2S3 remaining is 17.212 g

Explanation:

Equation of the reaction is given below:

Al2S3 + 6H2O > 2Al(OH)3 + 3H2S

From the balanced equation above

6 mole of H20 reacts with 1 mole of Al2S3

i.e. 6 * 18.02 g of H2O reacts with 1 * 150.71 g of Al2S3

= 108.12 g of H2O reacts with 150.71 g of Al2S3

Therefore 2.0 g of water will react with 2.0 * (150.71/108.12) g of Al2S3

= 2.788 g of Al2S3

Mass of Al2S3 remaining = 20.0 g - 2.788 g = 17.212 g

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.