Question

tu désires neutraliser une solution basique. pour ce faire, tu utilises 200 ml dune solution acide dont la concentration est de 5 g/l. quelle sera la quantité de mol de sel obtenu si les deux réactifs réagissent complètement? h₂s (aq) + naoh(aq) → h₂o (l) + na₂s (aq) équation non balancée 1 point a 0,015 mol b 15,000 mol c 0,030 mol d 3,000 mol e 1,000 mol f 1,500 mol g 0,06 mol

Explanation:

Step 1: Balance the chemical equation

The unbalanced equation is $\ce{H_{2}S_{(aq)} + NaOH_{(aq)} -> H_{2}O_{(l)} + Na_{2}S_{(aq)}}$.
To balance it, we need 2 moles of $\ce{NaOH}$ to react with 1 mole of $\ce{H_{2}S}$ to form 2 moles of $\ce{H_{2}O}$ and 1 mole of $\ce{Na_{2}S}$.
Balanced equation: $\ce{H_{2}S_{(aq)} + 2NaOH_{(aq)} -> 2H_{2}O_{(l)} + Na_{2}S_{(aq)}}$

Step 2: Calculate mass of $\ce{H_{2}S}$ used

Volume of acid solution = 200 mL = 0.2 L (since 1 L = 1000 mL).
Concentration of $\ce{H_{2}S}$ solution = 5 g/L.
Mass of $\ce{H_{2}S}$ = concentration × volume = $5\ \text{g/L} \times 0.2\ \text{L} = 1\ \text{g}$.

Step 3: Calculate moles of $\ce{H_{2}S}$

Molar mass of $\ce{H_{2}S}$: $M = 2(1) + 32 = 34\ \text{g/mol}$.
Moles of $\ce{H_{2}S}$, $n = \frac{\text{mass}}{\text{molar mass}} = \frac{1\ \text{g}}{34\ \text{g/mol}} \approx 0.0294\ \text{mol}$ (≈ 0.03 mol).

Step 4: Relate moles of $\ce{H_{2}S}$ to moles of $\ce{Na_{2}S}$ (salt)

From the balanced equation, 1 mole of $\ce{H_{2}S}$ produces 1 mole of $\ce{Na_{2}S}$.
Thus, moles of $\ce{Na_{2}S}$ = moles of $\ce{H_{2}S} \approx 0.03\ \text{mol}$.

Answer:

C. 0,030 mol