When 22.4 litres of H _ 2 (g) is mixed with 11.2 litres of C l _ 2 (g), each at STP, the moles of HCl(g) formed is equal to

Correct answer: A. 1molofHCl(g) cases H _ 2 _ 22.4lt + & C l _ 2 _ 11.2lt → & 2HCl cases cases 1mole & = (1)/(2)mole cases Limiting reagent is C l _ 2 . So, 1 mole HCl is formed.

NEET 2014

Exam: NEET
Year: 2014
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

NEET 2014 · Chemistry · Some Basic Concepts of Chemistry

When 22.4 litres of H _ 2 (g) is mixed with 11.2 litres of C l _ 2...

When

22.4

litres of

{H}_{2}\left(g\right)

is mixed with

11.2

litres of

C{l}_{2}\left(g\right),

each at STP, the moles of

HCl\left(g\right)

formed is equal to

Options

A $1molofHCl\left(g\right)$
B $2molofHCl\left(g\right)$
C0.5 $mol$ HCl\left(g\right)
D $1.5molofHCl\left(g\right)$

Correct answer

A. $1molofHCl\left(g\right)$

Step-by-step solution

\begin{cases}\mathop{{H}_{2}}_{22.4lt} + & \mathop{C{l}_{2}}_{11.2lt} \to & 2HCl\end{cases}

\begin{cases}1mole & = \dfrac{1}{2}mole\end{cases}

Limiting reagent is

C{l}_{2} .

So,

1

mole

HCl

is formed.

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