Question

50.0 kg of N, (g) and 10.0 kg of H, (g) are
mixed to produce NH, (g). Calculate the
amount of NH, (g) formed. Identify the limiting reagent in the production of NH3 in this situation

pls say me how to do limited reagent sum frnds..?? I want ur help
I want with steps which is easy to understand​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of Number of Moles has been used. We know that the limiting reagent is the reactant whose quantity is less than the required quantity. We can find this by comparing the quantity of the reactants. After that we can calculate the amount of Ammonia formed.

____________________________________________

★ Formula Used :-

$\\\;\boxed{\sf{No.\;of\;Moles\;=\;\bf{\dfrac{Given\;Mass}{Molar\;Mass}}}}$

____________________________________________

★ Solution :-

Given,

» Molecular Mass of N₂ = 2(14) = 28 g

» Given Mass of N₂ = 50 Kg = 50 × 10³ g

» Molecular Mass of H₂ = 2(1) = 2g

» Given Mass of H₂ = 10 Kg = 10 × 10³ g

» Molecular of Mass of NH₃ = 17 g

____________________________________________

~ For the No. of Moles of H₂ and N₂ ::

$\\\;\;\sf{:\rightarrow\;\;No.\;of\;Moles\;=\;\bf{\dfrac{Given\;Mass}{Molar\;Mass}}}$

$\\\;\;\sf{:\rightarrow\;\;No.\;of\;Moles\;of\;H_{2}\;=\;\bf{\dfrac{10\;\times\;10^{3}}{2}}}$

$\\\;\;\bf{:\rightarrow\;\;No.\;of\;Moles\;of\;H_{2}\;=\;\bf{\green{5\;\times\;10^{3}\;\;g}}}$

$\\\;\;\sf{:\rightarrow\;\;No.\;of\;Moles\;of\;N_{2}\;=\;\bf{\dfrac{50\;\times\;10^{3}}{28}}}$

$\\\;\;\bf{:\rightarrow\;\;No.\;of\;Moles\;of\;N_{2}\;=\;\bf{\green{17.86\;\times\;10^{2}\;\;g}}}$

____________________________________________

~ From number of moles required in reaction ::

The balanced chemical equation for this reaction is given as,

$\\\;\sf{\pink{N_{2\;(g)}\;+\;3H_{2\;(g)}\;\longrightarrow\;\;\bf{2NH_{3\;(g)}}}}$

From this reaction we see,

1 mol of N₂ requires 3 mol of H₂ to form 2 mol of NH₃

This means, number of moles of H₂ required to react with 17.86 × 10² moles of N₂ is given as,

$\\\;\sf{\Rightarrow\;\;No.\;of\;moles\;of\;H_{2}\;required\;=\;3\;\times\;17.86\;\times\;10^{2}}$

$\\\;\bf{\Rightarrow\;\;No.\;of\;moles\;of\;H_{2}\;required\;=\;\red{5.36\times\;10^{3}}}$

But from above results we see that we have we have only 5 × 10³ moles of H₂ to react. This means, Hydrogen is the limiting reagent.

$\\\;\underline{\boxed{\tt{Limiting\;Reagent\;=\;\purple{H_{2}}}}}$

____________________________________________

~ For amount of NH₃ formed ::

Again from the reaction we that, we need 2 moles of NH₃ to be formed from 3 moles H₂.

This means,

$\\\;\sf{\Longrightarrow\;\;1\;mole\;of\;NH_{3}\;=\;\dfrac{2}{3}\;\times\;mole\;of\;H_{2}}$

Now we already know the number of moles of H₂ formed. Then,

$\\\;\sf{\Longrightarrow\;\;Moles\;of\;NH_{3}\;formed\;=\;\dfrac{3}{2}\;\times\;(5\;\times\;10^{3})}$

$\\\;\bf{\Longrightarrow\;\;Moles\;of\;NH_{3}\;formed\;=\;\gray{3.33\;\times\;10^{3}}}$

Now from this we can calculate the mass of NH₃ formed, as

$\\\;\;\sf{:\rightarrow\;\;No.\;of\;Moles\;=\;\bf{\dfrac{Given\;Mass}{Molar\;Mass}}}$

$\\\;\;\sf{:\rightarrow\;\;Mass\;Formed\;=\;\bf{No.\;of\;Moles\;\times\;Molar\;Mass}}$

$\\\;\;\sf{:\rightarrow\;\;Mass\;Formed\;=\;\bf{17\;\times\;3.33\;\times\;10^{3}}}$

$\\\;\;\sf{:\rightarrow\;\;Mass\;Formed\;=\;\bf{56.61\;\times\;10^{3}\;\;g}}$

$\\\;\;\bf{:\rightarrow\;\;Mass\;Formed\;=\;\bf{\orange{56.61\;\;Kg}}}$

$\\\;\underline{\boxed{\tt{Amount\;\;of\;\;NH_{3}\;formed\;=\;\bf{\blue{56.61\;\;Kg}}}}}$

____________________________________________

★ More to know :-

$\\\;\sf{\leadsto\;\;Density\;=\;\dfrac{Mass}{Volume}}$

$\\\;\sf{\leadsto\;\;Planck's\;Constant\;=\;\dfrac{Energy}{Frequency}}$

$\\\;\sf{\leadsto\;\;Latent\;Heat\;=\;\dfrac{Heat}{Mass}}$

$\\\;\sf{\leadsto\;\;Entropy\;=\;\dfrac{Heat}{Temperature}}$

$\\\;\sf{\leadsto\;\;Boltzmann's\;Constant\;=\;\dfrac{Energy}{Temperature}}$

Answer 2

Explanation:

Here the concept of Number of Moles has been used. We know that the limiting reagent is the reactant whose quantity is less than the required quantity. We can find this by comparing the quantity of the reactants. After that we can calculate the amount of Ammonia formed.

____________________________________________

★ Formula Used :-

\begin{gathered}\\\;\boxed{\sf{No.\;of\;Moles\;=\;\bf{\dfrac{Given\;Mass}{Molar\;Mass}}}}\end{gathered}

No.ofMoles=

MolarMass

GivenMass⬇️⬇️⬇️

____________________________________________

★ Solution :-⤵️⤵️

Given,

» Molecular Mass of N₂ = 2(14) = 28 g

» Given Mass of N₂ = 50 Kg = 50 × 10³ g

» Molecular Mass of H₂ = 2(1) = 2g

» Given Mass of H₂ = 10 Kg = 10 × 10³ g

» Molecular of Mass of NH₃ = 17 g

____________________________________________

~ For the No. of Moles of H₂ and N₂ ::✔️✔️

\begin{gathered}\\\;\;\sf{:\rightarrow\;\;No.\;of\;Moles\;=\;\bf{\dfrac{Given\;Mass}{Molar\;Mass}}}\end{gathered}

:→No.ofMoles=

MolarMass

GivenMass

⬇️⬇️✌️✌️

\begin{gathered}\\\;\;\sf{:\rightarrow\;\;No.\;of\;Moles\;of\;H_{2}\;=\;\bf{\dfrac{10\;\times\;10^{3}}{2}}}\end{gathered}

:→No.ofMolesofH

2

=

2

10×10

3

\begin{gathered}\\\;\;\bf{:\rightarrow\;\;No.\;of\;Moles\;of\;H_{2}\;=\;\bf{\green{5\;\times\;10^{3}\;\;g}}}\end{gathered}

:→No.ofMolesofH

2

=5×10

3

g

\begin{gathered}\\\;\;\sf{:\rightarrow\;\;No.\;of\;Moles\;of\;N_{2}\;=\;\bf{\dfrac{50\;\times\;10^{3}}{28}}}\end{gathered}

:→No.ofMolesofN

2

=

28

50×10

3

\begin{gathered}\\\;\;\bf{:\rightarrow\;\;No.\;of\;Moles\;of\;N_{2}\;=\;\bf{\green{17.86\;\times\;10^{2}\;\;g}}}\end{gathered}

:→No.ofMolesofN

2

=17.86×10

2

g

____________________________________________

~ From number of moles required in reaction ::

The balanced chemical equation for this reaction is given as,

\begin{gathered}\\\;\sf{\pink{N_{2\;(g)}\;+\;3H_{2\;(g)}\;\longrightarrow\;\;\bf{2NH_{3\;(g)}}}}\end{gathered}

N

2(g)

+3H

2(g)

⟶2NH

3(g)

From this reaction we see,

1 mol of N₂ requires 3 mol of H₂ to form 2 mol of NH₃

This means, number of moles of H₂ required to react with 17.86 × 10² moles of N₂ is given as,

\begin{gathered}\\\;\sf{\Rightarrow\;\;No.\;of\;moles\;of\;H_{2}\;required\;=\;3\;\times\;17.86\;\times\;10^{2}}\end{gathered}

⇒No.ofmolesofH

2

required=3×17.86×10

2

\begin{gathered}\\\;\bf{\Rightarrow\;\;No.\;of\;moles\;of\;H_{2}\;required\;=\;\red{5.36\times\;10^{3}}}\end{gathered}

⇒No.ofmolesofH

2

required=5.36×10

3

But from above results we see that we have we have only 5 × 10³ moles of H₂ to react. This means, Hydrogen is the limiting reagent.

\begin{gathered}\\\;\underline{\boxed{\tt{Limiting\;Reagent\;=\;\purple{H_{2}}}}}\end{gathered}

LimitingReagent=H

2

____________________________________________

~ For amount of NH₃ formed ::

Again from the reaction we that, we need 2 moles of NH₃ to be formed from 3 moles H₂.

This means,

\begin{gathered}\\\;\sf{\Longrightarrow\;\;1\;mole\;of\;NH_{3}\;=\;\dfrac{2}{3}\;\times\;mole\;of\;H_{2}}\end{gathered}

⟹1moleofNH

3

=

3

2

×moleofH

2

Now we already know the number of moles of H₂ formed. Then,

\begin{gathered}\\\;\sf{\Longrightarrow\;\;Moles\;of\;NH_{3}\;formed\;=\;\dfrac{3}{2}\;\times\;(5\;\times\;10^{3})}\end{gathered}

⟹MolesofNH

3

formed=

2

3

×(5×10

3

)

\begin{gathered}\\\;\bf{\Longrightarrow\;\;Moles\;of\;NH_{3}\;formed\;=\;\gray{3.33\;\times\;10^{3}}}\end{gathered}

⟹MolesofNH

3

formed=3.33×10

3

Now from this we can calculate the mass of NH₃ formed, as

\begin{gathered}\\\;\;\sf{:\rightarrow\;\;No.\;of\;Moles\;=\;\bf{\dfrac{Given\;Mass}{Molar\;Mass}}}\end{gathered}

:→No.ofMoles=

MolarMass

GivenMass

\begin{gathered}\\\;\;\sf{:\rightarrow\;\;Mass\;Formed\;=\;\bf{No.\;of\;Moles\;\times\;Molar\;Mass}}\end{gathered}

:→MassFormed=No.ofMoles×MolarMass

\begin{gathered}\\\;\;\sf{:\rightarrow\;\;Mass\;Formed\;=\;\bf{17\;\times\;3.33\;\times\;10^{3}}}\end{gathered}

:→MassFormed=17×3.33×10

3

\begin{gathered}\\\;\;\sf{:\rightarrow\;\;Mass\;Formed\;=\;\bf{56.61\;\times\;10^{3}\;\;g}}\end{gathered}

:→MassFormed=56.61×10

3

g

\begin{gathered}\\\;\;\bf{:\rightarrow\;\;Mass\;Formed\;=\;\bf{\orange{56.61\;\;Kg}}}\end{gathered}

:→MassFormed=56.61Kg

\begin{gathered}\\\;\underline{\boxed{\tt{Amount\;\;of\;\;NH_{3}\;formed\;=\;\bf{\blue{56.61\;\;Kg}}}}}\end{gathered}

AmountofNH

3

formed=56.61Kg

____________________________________________

★ More to know :-

\begin{gathered}\\\;\sf{\leadsto\;\;Density\;=\;\dfrac{Mass}{Volume}}\end{gathered}

⇝Density=

Volume

Mass

\begin{gathered}\\\;\sf{\leadsto\;\;Planck's\;Constant\;=\;\dfrac{Energy}{Frequency}}\end{gathered}

⇝Planck

′

sConstant=

Frequency

Energy

\begin{gathered}\\\;\sf{\leadsto\;\;Latent\;Heat\;=\;\dfrac{Heat}{Mass}}\end{gathered}

⇝LatentHeat=

Mass

Heat

\begin{gathered}\\\;\sf{\leadsto\;\;Entropy\;=\;\dfrac{Heat}{Temperature}}\end{gathered}

⇝Entropy=

Temperature

Heat

\begin{gathered}\\\;\sf{\leadsto\;\;Boltzmann's\;Constant\;=\;\dfrac{Energy}{Temperature}}\end{gathered}

⇝Boltzmann

′

sConstant=

Temperature

Energy....