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Number of moles of $N{{H}_{3}}$ formed when 0.535g of $N{{H}_{4}}Cl$ is completely decomposed by NaOH, is:
\[N{{H}_{4}}Cl+NaOH\to N{{H}_{3}}+NaCl+{{H}_{2}}O\]
A. 0.01 mol
B. 5.35 mol
C. 1.7g
D. 0.17 mol


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Last updated date: 13th Jun 2024
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Answer
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Hint: To solve this question we will first find the molar mass of NaOH, and then will calculate the number of moles of NaOH. As we know that the number of moles is given by the formula =
\[Number\text{ }of\text{ }moles=\dfrac{molecular\text{ }weight}{molar\text{ }mass}\]

Complete Step by step solution:
- We can see in the reaction that $N{{H}_{4}}Cl$ is reacting with NaOH and sodium chloride and ammonia is formed as product:
\[N{{H}_{4}}Cl+NaOH\to N{{H}_{3}}+NaCl+{{H}_{2}}O\]
- We are being provided with the 0.535g of $N{{H}_{4}}Cl$. Now, firstly we will calculate the molar mass of NaOH as:
Molar mass=
\[\begin{align}
& NaOH \\
&\implies 22.9+16+1 \\
&\implies 39.9 \\
&\implies 40 \\
\end{align}\]
 - Hence, we can write that the molar mass of NaOH is 40 g.
- According to the reaction we can say that 53.5 g of$N{{H}_{4}}Cl$ will react completely with 40 g of NaOH.
Therefore, we can say that 0.535 g of $N{{H}_{4}}Cl$ will totally react with 0.4 g NaOH.
- As we know that the number of moles=
\[Number\text{ }of\text{ }moles=\dfrac{molecular\text{ }weight}{molar\text{ }mass}\]
Therefore, putting the values in above equation, we get:
Number of moles of NaOH =
$\begin{align}
& \dfrac{0.4}{40} \\
&\implies 0.01M \\
\end{align}$

- Hence, we conclude that the correct option is (a), that is 0.01 moles of $N{{H}_{3}}$ is formed when 0.535g of $N{{H}_{4}}Cl$ is completely decomposed by NaOH .

Note: - Mole is having SI unit of mol, or we can say that it is the SI unit of amount of substance. One mole is found to contain $6.022\times {{10}^{23}}$ elementary entities.
- We should not forget to write units after solving any question.