Question

The minimum quantity of H2S needed to precipitate 127 g of Ag+ will be nearly:(At. Mass of Ag=108]
A 120.5 g
B 20g
с 688
D 44 g
please tell the answer of this question...​

Answer 1

The minimum quantity of H2S needed to precipitate 127 g of Ag+ will be nearly  20 g.

Chemical Equation:

2Ag+  +  H2S  =Ag2S  + H2

• Atomic mass of Ag = 108 g

• Molecular weigth of H2S = 2+ 32= 34 g

       According to the equation:

• 2 x 108 g Ag requires = 34 g of H2S.
• 216 g of  Ag requires = 34 g of H2S.
• 127 g of  Ag requires = (34/216) x 127 g of H2S.

                                           = 19.9 g of H2S.

                                           =  20  g of H2S.