My best Chemistry question as yet. Have a look!

Let the amount of nitric acid in the mixture be $a$ grams and the amount of sulphuric acid in the mixture be $b$ grams.
Then milliequivalence of calcium hydroxide used is equal to the total milliequivalence of the contents of the mixture.

Milliequivalence of $HNO_3$ = $\dfrac{a}{\dfrac{63}{1}} × 1000 = \dfrac{1000a}{63}$

Milliequivalence of $H_2SO_4$ = $\dfrac{b}{\dfrac{98}{2}} × 1000 = \dfrac{1000a}{49}$

Milliequivalence of $Ca{(OH)}_2$ = $\dfrac{5.6 × 2.8}{0.3} = 52.27$

Therefore, $\dfrac{1000a}{63} + \dfrac{1000b}{49} = 52.27\\ \dfrac{7000a + 9000b}{441} = 52.27\\ 7000a + 9000b = 23051.57 \longrightarrow \boxed{1}$

Now, we know that the total volume of the mixture is $2.7$ g, so:-
$a + b = 2.7 \longrightarrow \boxed{2}$

Multiplying $\boxed{2}$ by $7000$ , we get:-
$7000a + 7000b = 18900 \longrightarrow \boxed{3}$

Subtracting $\boxed{3}$ from $\boxed{1}$ , we get:-
$2000b = 4151.07\\ b = 2.08$

So, % composition of sulphuric acid in mixture = $\dfrac{2.08}{2.7} × 100\\ = 77.04\\ \approx \boxed{77\%}$ .

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$\text{NOTE}$ :-
There are two formulas for calculating milliequivalence.
1) $\dfrac{\text{given weight} × \text{n-factor}}{\text{molecular weight}} × 100$
2) $\text{Normality} × \text{Volume} = \text{Normality} × \dfrac{\text{Mass}}{\text{Density}}$

I have used the former method for the calculation of milliequivalence of the contents of the mixture because normality was not mentioned. Whereas I have used the latter method in calculation of the milliequivalence of calcium hydroxide because that it easier when normality of the solution is mentioned.