
If 16.4 g of calcium nitrate is heated. Calculate the volume of \[N{{O}_{2}}\] obtained and mass of CaO obtained.
\[2Ca{{(N{{O}_{3}})}_{2}}\to 2CaO+4N{{O}_{2}}+{{O}_{2}}\]
Answer
574.8k+ views
Hint: Calcium nitrate decomposes and forms two moles of calcium oxide, four moles of nitrogen dioxide and one mole of oxygen. Calcium oxide (CaO) is a solid, nitrogen dioxide (\[N{{O}_{2}}\]) as a liquid when cooled and oxygen (\[{{O}_{2}}\]) is a gaseous substance formed when calcium nitrates decompose.
Complete step by step solution:
In the question, it is given that 16.4 gm of calcium nitrate is heated.
The given chemical reaction is as follows.
\[2Ca{{(N{{O}_{3}})}_{2}}\to 2CaO+4N{{O}_{2}}+{{O}_{2}}\]
Molar mass of \[Ca{{(N{{O}_{3}})}_{2}}\]= 40 + 2 (14) + 6 (16) = 164.
Number of moles of
\[\begin{align}
& Ca{{(N{{O}_{3}})}_{2}}=\dfrac{given\text{ }weight}{molecular\text{ }weight\text{ }of\text{ }the\text{ }subs\tan ce} \\
& \text{ = }\dfrac{16.4}{164}=0.1 \\
\end{align}\]
From the given equation we can say that 2 moles \[Ca{{(N{{O}_{3}})}_{2}}\]gives 4 moles of \[N{{O}_{2}}\].
Therefore 0.1 moles of \[Ca{{(N{{O}_{3}})}_{2}}\]is undergoing decomposition and gives 0.2 moles of \[N{{O}_{2}}\].
Volume of \[N{{O}_{2}}\]at STP
\[\begin{align}
& \text{=number of moles of N}{{\text{O}}_{\text{2}}}\text{ }\!\!\times\!\!\text{ volume in liter} \\
& \text{=0}\text{.2}\times \text{22}\text{.4} \\
& \text{= 4}\text{.48L} \\
\end{align}\]
2 moles of \[Ca{{(N{{O}_{3}})}_{2}}\] gives 2 moles of CaO (from the given chemical reaction).
Then 0.1 moles of \[Ca{{(N{{O}_{3}})}_{2}}\]gives 0.1 moles of CaO.
Therefore mass of Calcium oxide (CaO)
\[\begin{align}
& \text{= number of moles of CaO }\!\!\times\!\!\text{ molecular weight of CaO} \\
& \text{= 0}\text{.1}\times \text{56} \\
& \text{=5}\text{.6g} \\
\end{align}\]
Therefore 16.4 g of calcium nitrate decomposes and forms 4.48 litre of \[N{{O}_{2}}\] and 16.4 g of calcium nitrate decomposes and forms 5.6 g of Calcium oxide (CaO).
Note: Calcium nitrate decomposes and forms nitrogen oxides later those nitrogen oxides can be converted to nitric acid. Calcium nitrate is mostly produced as a fertilizer and used as nutrition for plants and in the treatment of wastewater. Calcium nitrate is a good source of calcium and nitrogen for plants.
Complete step by step solution:
In the question, it is given that 16.4 gm of calcium nitrate is heated.
The given chemical reaction is as follows.
\[2Ca{{(N{{O}_{3}})}_{2}}\to 2CaO+4N{{O}_{2}}+{{O}_{2}}\]
Molar mass of \[Ca{{(N{{O}_{3}})}_{2}}\]= 40 + 2 (14) + 6 (16) = 164.
Number of moles of
\[\begin{align}
& Ca{{(N{{O}_{3}})}_{2}}=\dfrac{given\text{ }weight}{molecular\text{ }weight\text{ }of\text{ }the\text{ }subs\tan ce} \\
& \text{ = }\dfrac{16.4}{164}=0.1 \\
\end{align}\]
From the given equation we can say that 2 moles \[Ca{{(N{{O}_{3}})}_{2}}\]gives 4 moles of \[N{{O}_{2}}\].
Therefore 0.1 moles of \[Ca{{(N{{O}_{3}})}_{2}}\]is undergoing decomposition and gives 0.2 moles of \[N{{O}_{2}}\].
Volume of \[N{{O}_{2}}\]at STP
\[\begin{align}
& \text{=number of moles of N}{{\text{O}}_{\text{2}}}\text{ }\!\!\times\!\!\text{ volume in liter} \\
& \text{=0}\text{.2}\times \text{22}\text{.4} \\
& \text{= 4}\text{.48L} \\
\end{align}\]
2 moles of \[Ca{{(N{{O}_{3}})}_{2}}\] gives 2 moles of CaO (from the given chemical reaction).
Then 0.1 moles of \[Ca{{(N{{O}_{3}})}_{2}}\]gives 0.1 moles of CaO.
Therefore mass of Calcium oxide (CaO)
\[\begin{align}
& \text{= number of moles of CaO }\!\!\times\!\!\text{ molecular weight of CaO} \\
& \text{= 0}\text{.1}\times \text{56} \\
& \text{=5}\text{.6g} \\
\end{align}\]
Therefore 16.4 g of calcium nitrate decomposes and forms 4.48 litre of \[N{{O}_{2}}\] and 16.4 g of calcium nitrate decomposes and forms 5.6 g of Calcium oxide (CaO).
Note: Calcium nitrate decomposes and forms nitrogen oxides later those nitrogen oxides can be converted to nitric acid. Calcium nitrate is mostly produced as a fertilizer and used as nutrition for plants and in the treatment of wastewater. Calcium nitrate is a good source of calcium and nitrogen for plants.
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