Question

11.2 ml Of Sulphur dioxide at NTP in weight in gram​

Answer 1

Answer:

we have, 11.2 L of SO 2

= 0.5 mole

for 1 mole of SO 2

we require 2 moles of H 2

S

and for 0.5 mole of SO 2

we require 1 moles of H 2

S

so, we have SO 2

as a limiting reagent

1 mole of SO 2

form 3 mole of sulphur

0.5 mole of SO 2

produce S = 0.5 * 3 = 1.5 mole

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Answer 2

Step-by-step explanation:

we have, 11.2 L of S02

= 0.5 mole

for 1 mole of so2

we require 2 moles of H2S

and for 0.5 mole of SO2

we require 1 moles of H2s

so, we have SO2 as a limiting reagent 1 mole of SO2 form 3 mole of sulphur0.5 mole of SO2 produce S = 0.5 * 3 = 1.5 mol