Question

How many liters of a 0.75 M solution of $Ca{{(N{{O}_{3}})}_{2}}$ will be required to react with 148 g of $N{{a}_{2}}C{{O}_{3}}$? $\_\text{ }Ca{{\left( N{{O}_{3}} \right)}_{2}}\text{ }+\text{ }\_\text{ }N{{a}_{2}}C{{O}_{3}}~\to \_\text{ }CaC{{O}_{3}}\text{ }+\text{ }\_\text{ }NaN{{O}_{3}}$

Answer 1

Hint: First we will balance the given chemical equation. Then further calculate the moles of $N{{a}_{2}}C{{O}_{3}}$ and $Ca{{(N{{O}_{3}})}_{2}}$ followed by calculating the volume of 0.75 M solution of $Ca{{(N{{O}_{3}})}_{2}}$, that will be required to react with 148 g of $N{{a}_{2}}C{{O}_{3}}$.

Formula used: The formulas required for solving this question are as follows:-
$n=\dfrac{w}{M}$

Complete step by step answer:
-We will begin with balancing of the given chemical equation:-
$Ca{{\left( N{{O}_{3}} \right)}_{2}}\text{ }+\text{ }N{{a}_{2}}C{{O}_{3}}~\to CaC{{O}_{3}}\text{ }+\text{ }NaN{{O}_{3}}$

Number of atoms in reactant side:	Number of atoms in product side:
Ca = 1	Ca = 1
N = 2	N = 1
O = 9	O = 6
Na = 2	Na = 1
C = 1	C = 1

As we can see, the number of Ca, Na and C atoms are the same on both sides whereas the number of N and O atoms is different. Therefore we will multiply $NaN{{O}_{3}}$ with 2 to bring increment in the number of N atoms.
Now the equation is:-
$Ca{{\left( N{{O}_{3}} \right)}_{2}}\text{ }+\text{ }N{{a}_{2}}C{{O}_{3}}~\to CaC{{O}_{3}}\text{ }+\text{ 2 }NaN{{O}_{3}}$

Number of atoms in reactant side:	Number of atoms in product side:
Ca = 1	Ca = 1
N = 2	N = 2
O = 9	O = 9
Na = 2	Na = 1
C = 1	C = 1

As we can see, the number of Ca, Na, O, N and C atoms are same on both sides of the equation. Hence this is the balanced chemical equation.

-Calculation of number of moles $N{{a}_{2}}C{{O}_{3}}$ of:-
$n=\dfrac{w}{M}$
Where,
n= Number of moles
w = given mass of the substance or compound =148g of $N{{a}_{2}}C{{O}_{3}}$
M= molar mass of the substance or compound ($N{{a}_{2}}C{{O}_{3}}$) =$23\times 2+12+16\times 3=106g$
Number of moles of $N{{a}_{2}}C{{O}_{3}}$= $n=\dfrac{w}{M}$:
$\Rightarrow n=\dfrac{148g}{106g}=1.396moles$

-Calculation of the volume of 0.75 M solution of $Ca{{(N{{O}_{3}})}_{2}}$that will be required to react with 148 g of$N{{a}_{2}}C{{O}_{3}}$
$Ca{{\left( N{{O}_{3}} \right)}_{2}}\text{ }+\text{ }N{{a}_{2}}C{{O}_{3}}~\to CaC{{O}_{3}}\text{ }+\text{ 2 }NaN{{O}_{3}}$
From the above reaction, we can see that 1 mole reacts with 1 mole of$Ca{{(N{{O}_{3}})}_{2}}$ . Then, 1.396 moles of react with = $1.396\text{ moles of N}{{\text{a}}_{2}}C{{O}_{3}}\times \dfrac{1\text{mole of }Ca{{(N{{O}_{3}})}_{2}}}{1\text{mole of N}{{\text{a}}_{2}}C{{O}_{3}}}=1.396\text{ moles of }Ca{{(N{{O}_{3}})}_{2}}$
As we have been given that 0.75M solution of $Ca{{(N{{O}_{3}})}_{2}}$which means 0.75 moles of $Ca{{(N{{O}_{3}})}_{2}}$ present in 1L of solution.
Therefore, 1.396 moles of $Ca{{(N{{O}_{3}})}_{2}}$will be present in = $1.396\text{ moles of }Ca{{(N{{O}_{3}})}_{2}}\times \dfrac{1L}{\text{0}\text{.75mole of }Ca{{(N{{O}_{3}})}_{2}}}=1.86L\text{ of solution}$
 -Hence, 1.86L of 0.75 M solution of $Ca{{(N{{O}_{3}})}_{2}}$will be required to react with 148 g of $N{{a}_{2}}C{{O}_{3}}$.

Note: -Molarity is defined as the number of moles of solute present in 1L of solution.
 - Whenever we multiply atoms with some number, always write down the no. of atoms on both sides because we multiply the number with not only that particular atom but with all the atoms that the molecule is composed of.
-Kindly prefer to note down all the values and convert them into required units so as to obtain the desired results accurately.