Question

0.01 mole of chl3 reacts with ag to produce a gas at ntp whose volume is
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Answer 1

Explanation:

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Answer 2

$\checkmark$ Answer:

Volume = 112 ml

$\checkmark$ Explanation:

$\star$ Chemical Equation:

$\red{\boxed{\sf 2CHI_{3}+6Ag \longrightarrow C_{2}H_{2}+6AgI}}$

We can understand that,

2 moles of Iodoform produces 1 mole of Acetylene

Mathematically,

$\sf 2 \ mole \longrightarrow 1 \ mole$

Given that,

0.01 mole of CHI₃ reacts with Ag

Hence,

$\sf 0.01 \ mole \longrightarrow x \ moles$

Therefore,

We need to find "x"

Hence,

$\implies \sf x=0.01 \times \dfrac{1}{2}$

$\green{\sf \implies x=0.005 \ moles}$

Therefore,

0.01 mole of CHI₃ produces 0.005 mole of Acetylene

Given that,

0.01 mole of CHI₃ reacts with Ag to produce a gas at NTP

Condition = NTP

Hence,

The volume occupied by 0.005 moles of acetylene is

$\implies \sf Volume=0.005 \times 22.4 \ L$

$\implies \sf Volume=0.112 \ L$

$\blue{\boxed{\sf Volume=112 \ ml}}$