Question

The following chemical equation is given
$2Ca{\left( {N{O_3}} \right)_2}\xrightarrow{{}}2CaO + 4N{O_2} + {O_2}$
The molecular mass of calcium nitrate is $164u$, calculate the weight of calcium oxide on heating $16.4g$ of calcium nitrate . Also calculate the volume of nitrogen dioxide at standard temperature and pressure.

Answer 1

Hint: We are given chemical equation $2Ca{\left( {N{O_3}} \right)_2}\xrightarrow{{}}2CaO + 4N{O_2} + {O_2}$. From this equation we see that $2$moles of calcium oxide reacts with $2$ moles calcium nitrate and $4$moles of nitrogen dioxide. Atomic weights of $Ca,N,O$ are given , from this we can calculate the molar mass . The molar volume of gaseous compound is $22.4L$

Complete step by step solution:
We are given equation $2Ca{\left( {N{O_3}} \right)_2}\xrightarrow{{}}2CaO + 4N{O_2} + {O_2}$
According to the following equation, $2$moles of calcium nitrate on heating gives $2$ moles of calcium oxide , $4$ moles of nitrogen dioxide gas and $1$ mole of oxygen gas. Moles represents $6.022 \times {10^{23}}$ atoms, ions or molecules of that substance.
Now we will calculate the formula mass for the given equation. Formula mass is the sum of all the atomic weights in the empirical formula.
Formula mass for $2Ca{\left( {N{O_3}} \right)_2}$ $ = $ $2 \times \left[ {40 + 2\left( {14 + 48} \right)} \right]$
\[ \Rightarrow 2 \times \left[ {40 + 2\left( {14 + 48} \right)} \right] = 328amu\]
Formula mass for $2CaO$ $ = $ $2 \times \left[ {40 + 16} \right]$
\[ \Rightarrow 2 \times \left[ {40 + 16} \right] = 112amu\]
Nitrogen dioxide and oxygen are gaseous compounds, so we will multiply stoichiometric coefficient of $N{O_2}$by $22.4L$,the volume occupied by $1$ mole of gaseous substance at standard temperature and pressure.
Formula mass for $4N{O_2}$ $ = $ $4 \times 22.4L$
\[ \Rightarrow 4 \times 22.4L = 89.6L\]
Formula mass for ${O_2}$ $ = $ $1 \times 22.4L$
$ \Rightarrow 1 \times 22.4L = 22.4L$
Now we see that $328g$ of $Ca{\left( {N{O_3}} \right)_2}$ gives $112g$of $CaO$. To find how much $CaO$ is produced on heating $16.4g$ of $Ca{\left( {N{O_3}} \right)_2}$ , we apply a unitary method.
$1g$ of $Ca{\left( {N{O_3}} \right)_2}$$ \to \dfrac{{112}}{{328}}g$ of $CaO$
$16.4g$ of $Ca{\left( {N{O_3}} \right)_2}$$ \to \dfrac{{112}}{{328}} \times 16.4g$of $CaO$
$ \Rightarrow \;\dfrac{{112}}{{328}} \times 16.4$$ = 5.6g$

Moles is defined as the ratio of given mass to molar mass
Moles of $Ca{\left( {N{O_3}} \right)_2}$ $ = \dfrac{m}{M}$
$m = $Given mass
$M = $Molar mass
$ \Rightarrow \dfrac{m}{M} = \dfrac{{16.4}}{
  164 \\
    \\
 }$$ = 0.1M$
Applying unitary method again
Moles of $N{O_2}$ formed $ = \dfrac{4}{2} \times 0.1$
$ \Rightarrow \dfrac{4}{2} \times 0.1 = 0.2M$
Volume of $1$ mole of gas is $22.4L$
Volume of $0.2M$ of $N{O_2}$ $ = 0.2 \times 22.4L$
$ \Rightarrow 0.2 \times 22.4L = 4.48L$
Therefore Volume of $0.2M$ of $N{O_2}$ $= 4.48L$

Note: We should take note of the stoichiometric coefficients in the given equation. Also the units of measurements should be noted. There is an alternative method to calculate the mass of $CaO$ .
Moles of $Ca{\left( {N{O_3}} \right)_2}$ $ = 0.1M$
Moles of $Ca{\left( {N{O_3}} \right)_2}$ $ = $ Moles of $CaO$
Mass of $CaO$ $ = $ $0.1 \times 56 = 5.6g$