Hydrogen sulfide reacts with sulfur dioxide to give $H_{2} O$ and $S, H_{2} S + {SO}_{2} = H_{2} O + S (solid),$ unbalanced. If $3.0 L$ of $H_{2} S$ gas at $760$ torrs produced $4.8 g$ of sulfur, what is the temperature in Celsius?

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Solution

Step 1: Analyzing the given data:
Hydrogen sulfide reacts with sulphur dioxide to form water and Sulphur.
$\underset{Hydrogen sulfide}{H_{2} S (g)} + \underset{Sulphur dioxide}{{SO}_{2} (g)} \to \underset{Water}{H_{2} O (l)} + \underset{Sulphur}{S (s)}$
Also, $3.0 L of H_{2} S$ gas at $760 torr$ produced $4.8 g of Sulfur .$
Step 2: Calculating the number of moles of H₂S
The balanced chemical equation can be represented as:
$\underset{Hydrogen sulfide}{2 H_{2} S (g)} + \underset{Sulphuric acid}{{SO}_{2} (g)} \to \underset{Water}{2 H_{2} O (l)} + \underset{Sulphur}{3 S (s)}$
From stoichiometry,
3 moles of Sulphur requires $= 2 moles of H_{2} S$
$3 \times 32 g$ of Sulphur requires $= 2 \times 34 g of H_{2} S$
$1 g$ of Sulphur requires $= \frac{2 \times 34 g}{3 \times 32 g} of H_{2} S$
4.8 g of Sulphur requires $= \frac{2 \times 34 g}{3 \times 32 g} \times 4.8 g of H_{2} S$
Moles of $H_{2} S$ $= \frac{2 \times 34 g}{3 \times 32 g} \times \frac{4.8 g}{34 g {mol}^{- 1}} = 0.0998 mol$
Step 3: Calculating the temperature
Applying the Ideal gas law we know that;
$PV = nRT T = \frac{PV}{nR} T = \frac{1 atm \times 3.0 L}{0.0998 mol \times 0.0826 {LatmK}^{- 1} {mol}^{- 1}} = 366 K T = (366 - 273.15) º C T = 90 º C$
Hence, the temperature in celsius is calculated as $90 º C$ .

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