Question

(i) How many moles of sulphur will be produced when 2 moles of H2S reacts with 11.2L of SO2 at NTP.(ii) Name the limiting reagent in the above reaction.

Answer 1

Answer:

i) Solution

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Correct option is

A

1.5

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

Explanation:

ii) Solution

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24g

2mol

2C(s)

+

32g

1mol

O

2

→

56g

2mol

2CO(g)

Let carbon be completely consumed.

24g carbon give 56 g CO.

Let O

2

is completely consumed.

∵ 32 g O

2

give 56 g CO.

∴ 96 g O

2

Will give

32

56

×96gCO=168gCO

Since, carbon gives least amount of product, te.,56 g CO or 2 mole CO, hence carbon will be the limiting reactant.

∴ Excess reactant is O

2

.

Amount of O

2

used =56−24=32g

Amount of O

2

left =96−32=64g

32g O

2

react with 24 g carbon

∴ 96 g O

2

will react with 72g carbon.

Thus, carbon should be taken 72g so that nothing is left at the end of the reaction.