Question

The equation for the action of heat on calcium nitrate is:
$2Ca{{(N{{O}_{3}})}_{2}}\to 2CaO+4N{{O}_{2}}\uparrow +{{O}_{2}}\uparrow$
i.) How many moles of $N{{O}_{2}}$ are produced when 1 mole of $Ca{{(N{{O}_{3}})}_{2}}$ decomposes?
ii.) What volume of ${{O}_{2}}$ at S.T.P. will be produced on heating 65.6 g of $Ca{{(N{{O}_{3}})}_{2}}$?
iii.) Find out the mass of CaO formed when 65.6 g of $Ca{{(N{{O}_{3}})}_{2}}$, is heated.
iv.) Find out the mass of $Ca{{(N{{O}_{3}})}_{2}}$ required to produce 5 moles of gaseous products.
v.) Find out the mass of $Ca{{(N{{O}_{3}})}_{2}}$ required to produce 44.8 of $N{{O}_{2}}$ at S.T.P. (Relative molecular mass of $Ca{{(N{{O}_{3}})}_{2}}$ = 164 and of CaO = 56) on the ranic compound prepared by each of the following reactions.

Answer 1

To solve this question, we should have prior knowledge about the calculation of number of moles. Number of moles is defined as the ratio of given mass to the molar mass.

Complete Solution :
(i) The chemical reaction involved in this question is mentioned below:
$2Ca{{(N{{O}_{3}})}_{2}}\to 2CaO+4N{{O}_{2}}\uparrow +{{O}_{2}}\uparrow$
From the given reaction, we can observe that 2 moles of $Ca{{(N{{O}_{3}})}_{2}}$ produces 4 moles of $N{{O}_{2}}$ . So, one mole of $Ca{{(N{{O}_{3}})}_{2}}$ will produce $\dfrac{4}{2}$ mole of $N{{O}_{2}}$= 2 moles of $N{{O}_{2}}$ is produced.

(ii) Molar mass of $Ca{{(N{{O}_{3}})}_{2}}$ = 164 gm
In this equation 2 moles of $Ca{{(N{{O}_{3}})}_{2}}$ is produced, molar mass of 2 moles of $Ca{{(N{{O}_{3}})}_{2}}$ is $= 2\times 164g$ .
$2\times 164g$ of $Ca{{(N{{O}_{3}})}_{2}}$ produces 22.4 L of ${{O}_{2}}$
1g of $Ca{{(N{{O}_{3}})}_{2}}$produces $= \dfrac{22.4}{2\times 164}L$
65.6 g of $Ca{{(N{{O}_{3}})}_{2}}$produce =$\dfrac{22.4}{2\times 164}\times \dfrac{656}{10}=4.48L$ of oxygen is produced.

(iii) 328 g of $Ca{{(N{{O}_{3}})}_{2}}$ produce 112g of CaO
1 g of $Ca{{(N{{O}_{3}})}_{2}}$produce $\dfrac{112}{2\times 164}g$ of CaO
65.6 g of $Ca{{(N{{O}_{3}})}_{2}}$ will produce $=\dfrac{112}{328}\times \dfrac{656}{10} = 22.4g$
(iv)gaseous molecules are produced in the reaction = 4 moles of $N{{O}_{2}}$, and 1 mole of ${{O}_{2}}$ = 5 moles
Mass of $Ca{{(N{{O}_{3}})}_{2}}$ required will be the mass of 2 moles = $2\times 164 = 328g$

(v) $4\times 22.4L$ of $N{{O}_{2}}$ produce by 328 g of $Ca{{(N{{O}_{3}})}_{2}}$
1L of $N{{O}_{2}}$ will produce $= \dfrac{328}{4\times 22.4}$
44.8L of $N{{O}_{2}}$ will produce $\dfrac{328}{4\times 22.4}\times 44.8 = 164g$ of $Ca{{(N{{O}_{3}})}_{2}}$

Note: To solve this type of question, the most important part is to balance the chemical reaction because if the chemical reaction is not balanced then all the stoichiometric calculation is not possible. A chemical equation is balanced when the number of atoms present at the reactant side is equal to the number of atoms present at the product side.