A mixture of NaOH and NaCl weighs 2.0g and is dissolved in 250 cm3 of water. A 25.0 cm3 aliquot is titrated with 0.1M H2SO4, requiring 20.0 cm3. Calculate moles of H2SO4 and moles of NaOH.

The number of moles of sulphuric (VI) acid used is $\boxed{0.002 \text{ mol}}$.

b) c) Calculate the number of moles of the substance that reacted with sulphuric (VI) acid.
The substance that reacted with sulfuric acid is $\text{NaOH}$.
From the balanced equation, 1 mole of $\text{H}_2\text{SO}_4$ reacts with 2 moles of $\text{NaOH}$.

The number of moles of $\text{NaOH}$ that reacted is $\boxed{0.004 \text{ mol}}$.

b) d) Calculate the number of moles of the substance that would react with sulphuric (VI) acid in one litre of solution.
The 0.004 mol of $\text{NaOH}$ was present in a 25.0 cm$^3$ aliquot. The original solution had a total volume of 250 cm$^3$.

The number of moles of $\text{NaOH}$ in one litre of solution is $\boxed{0.160 \text{ mol}}$.

b) e) Determine the mass of the unreacted substance in the mixture of solution and hence calculate its percentage in the mixture.
The mixture contains $\text{NaOH}$ (reacted substance) and $\text{NaCl}$ (unreacted substance).

Molar mass of $\text{NaOH}$ = 23.0 (Na) + 16.0 (O) + 1.0 (H) = 40.0 g/mol.
Mass of $\text{NaOH}$ = Moles $\times$ Molar mass = 0.040 mol $\times$ 40.0 g/mol = 1.6 g.

Total mass of mixture = 2.0 g.
Mass of $\text{NaCl}$ = Total mass of mixture - Mass of $\text{NaOH}$

The mass of the unreacted substance ($\text{NaCl}$) is $\boxed{0.4 \text{ g}}$.
Its percentage in the mixture is $\boxed{20\%}$.

Question 3: (Combustion Analysis of a Hydrocarbon)

Given:

• Mass of $\text