Question

Hydrogen chloride gas (7.3 g) is in a container at a pressure of 2.5 atm and a temperature of 300 K.

(1.) How many moles of gas are present? 
(2.) How many molecules of gas are present? 
(3.) If all the gas were dissolved in 100 cm³ of water, what mass of silver nitrate would need to be added to precipitate all the chloride ion produced.




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Answer 1

QUESTION:-

Hydrogen chloride gas (7.3 g) is in a container at a pressure of 2.5 atm and a temperature of 300 K.

(1.) How many moles of gas are present?  

(2.) How many molecules of gas are present?  

(3.) If all the gas were dissolved in 100 cm³ of water, what mass of silver nitrate would need to be added to precipitate all the chloride ion produced.

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EXPLANATION:-

(i) How many moles of gas is present ?

We know that-;

$\longrightarrow \boxed{\bigstar\bf\;Moles=\frac{mass}{molar\;mass}\bigstar}$

In the given question-:

Moles =??

Mass=7.3 g        (GIVEN)

Molar mass= 36.46 g/mol

Putting value in the formula we get -:

$\rightarrow\ \bf\;Moles=\frac{3.6}{36.46}\\\\\\\rightarrow \bf\;Moles=\frac{730}{3646} \; \\\\\ \longrightarrow \underline{\bf\;0.20\;Mol}$                [Kindly refer 1st attachment for the divison]

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(ii)How many molecules of gas are present?

We know that-:

$\longrightarrow \boxed{\bigstar\bf\;Molecules=Moles\times Avogadro's\; constant(6.022\times10^{23)}\bigstar}$

Molecules= ??

Moles-0.20

Arogardo's constant=6.022×10²³

Putting value-:

$\rightarrow \bf\;Molecules=0.20\times6.022\times10^{23}\\\\\\\rightarrow \bf\;Molecules=1.2044\times10^{23}\\\\\\\longrightarrow \underline{\bf\;Molecules=1.204\times10^{23}}$

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(3) If all the gas were dissolved in 100 cm³ of water, what mass of silver nitrate would need to be added to precipitate all the chloride ion produced.

To find this first we will balance the equation of the reaction -:

AgNO₃+HCL→AgCl+HNO₃

Observing reaction we get-:

Moles of AgNO₃=Moles of HCL

Moles of AgNO₃=0.20 Mol   (Calculated above)

Now-:

$\longrightarrow \boxed{\bigstar\bf\;Moles=\frac{mass}{molar\;mass}\bigstar}$

Moles=0.20

Mass=??

Molar mass= 169.91

Putting values we get-:

$\rightarrow \bf\;0.20=\frac{mass}{169.91}\\\\\\\rightarrow \bf\;Mass=169.91\times0.20\\\\\\\longrightarrow \underline{\bf\;Mass=33.982\;g}$

Therefore of mass=33.982 or mass≈34 g

[REFER 2ND ATTACHMENT FOR MULTIPLICATION]

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