Question

\[100ml\] of $0.06M$ $Ca{(N{O_3})_2}$ is added to $50ml$ of $0.06M$ $N{a_2}{C_2}{O_4}$ . After the reaction is complete:
A.)$0.003$ moles of calcium oxalate will get precipitated.
B.) $0.003M$of excess $C{a^{2 + }}$ will remain in excess.
C.) $N{a_2}{C_2}{O_4}$ is the limiting reagent.
D.) $Ca{(N{O_3})_2}$ is the excess reagent.

Answer 1

Hint: The chemical reaction required to solve this question is:
$Ca{(N{O_3})_2} + N{a_2}{C_2}{O_4} \to Ca{C_2}{O_4} + 2NaN{O_3}$ .
Keep in note the coefficients of the reactants and the products formed.

Complete step by step answer:
Now if we talking in millimoles, from the equation: $Ca{(N{O_3})_2} + N{a_2}{C_2}{O_4} \to Ca{C_2}{O_4} + 2NaN{O_3}$
then on calculating, we get to know that-
For $Ca{(N{O_3})_2}$ there are $100 \times 0.06 = 6mmol$
For $N{a_2}{C_2}{O_4}$ there are $50 \times 0.6 = 3mmol$ which is also equal to $0.003$ moles.
Since $N{a_2}{C_2}{O_4}$ has less millimoles and thus it is the limiting reagent.
Now on completing the reaction we will be left with-
$N{a_2}{C_2}{O_4}$-> $0mmol$ (since it is the limiting reagent , it will get precipitated , thus making our option-A ,C and D correct.)
$Ca{(N{O_3})_2}$-> $6 - 3 = 3mmol$
$Ca{C_2}{O_4}$-> $3mmol$ since $N{a_2}{C_2}{O_4}$ was $3mmol$ initially.
$NaN{O_3}$-> since it is multiplied by $2$ in the equation it will generate $3 \times 2 = 6mmol$.
Now to check option B we will calculate the molarity of $C{a^{2 + }}$.
$C{a^{2 + }}$-> $\dfrac{{3 mmol}}{{(100 + 50)ml}} = \dfrac{3}{{150}} = 0.02M$ .
Thus option-B is wrong and option-A,C,D are correct.

Note:
The limiting reagent in a chemical reaction is a reactant that is totally consumed when the chemical reaction is completed. The amount of product formed is limited by this reagent, since the reaction cannot continue without it.
That’s how we calculated the moles of the products formed during the above equation.