Question

7) According to the following reaction, how many grams of sulfur are formed when 37.4 g of water are formed? 2 H2S(g) + SO2(g) - 3 S(s) + 2H20(1) A) 99.8 g S B) 66.6 g S C) 56.1 gs D) 44.4 g S E) 14.0 gs

Answer 1

Option (A) is correct.

99.8 grams of S formed.

solution:- reaction :- 2 H₂S (g) + SO₂ (g) 35 (s) + 2 H₂OUS First calculate : no. of moles of water formed:- mass of weeter formed = 37.4g. molar mass of water = 18.015 g/mol So, no. of moles of water formed = mass of water formed molar mass of water 37.4 g - 2.07605 mol 18.015 g/mol According to the given reaction - - 2 moles of H₂O(l) and 3 mol of Scs) formed, so, when 2.07605 mol of H₂O(l) formed no. of moles of 'Sess formed 13 x 2.07605 mol sis) = 3.114075 mol of Srs) formed molar mass of 's' (sulfur. ) = 32.06 g mol mass of 's formed = no. of moles of 's' formed x molar mass of 's = 3.114075 mol X 32.06 g/mol - = 99.84 grams of 's'. Igg. 84 grams of 's formed

Answer 2

Answer 3

Balanced chemical equation is:

2 H2S(g) + SO2(g) → 3 S(s) + 2 H2O(l)

No. of moles of water formed = (37.4 g) / (18.02 g/mol) = 2.075 mol

No. of moles of Sulfur formed = (3 / 2) x (2.075 mol) = 3.1 mol

Weight of Sulfur formed = (3.1 mol) x (32.06 g/mol) = 99.8 grams of Sulfur formed.