Under conditions of $NTP$ $"NTP"$ , what volume of $S O_{2} (g)$ $SO_2(g)$ results from burning a $10 \cdot g$ $10*g$ mass of elemental sulfur?

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Under conditions of $NTP$ $"NTP"$ , what volume of $S O_{2} (g)$ $SO_2(g)$ results from burning a $10 \cdot g$ $10*g$ mass of elemental sulfur?

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Answer 1 · 2017-05-19T05:45:29

We address the stoichiometric equation...........and I get a volume of under $10 \cdot L$ $10*L$ ...............

Explanation:

$S (s) + O_{2} (g) \to S O_{2} (g)$ $S(s)+O_2(g) rarrSO_2(g)$

Now depending on your definition of $NTP$ $"NTP"$ (they seem to change with the wind across syllabuses) we can work out the volume. I use values of $NTP$ $"NTP"$ to specify $P = 1 \cdot a t m$ $P=1*atm$ , and $T = 293.15 \cdot K$ $T=293.15*K$ .

$Moles of sulfur \equiv Moles of sulfur dioxide$ $"Moles of sulfur "-=" Moles of sulfur dioxide"$

$Moles of sulfur \equiv \frac{10 \cdot g}{32.06 \cdot g \cdot m o l^{- 1}} = 0.312 \cdot m o l$ $"Moles of sulfur"-=(10*g)/(32.06*g*mol^-1)=0.312*mol$ .

And thus there are $0.312 \cdot m o l$ $0.312*mol$ $S O_{2}$ $SO_2$ gas........given conversion stoichiometric conversion.

$V = \frac{n R T}{P} = \frac{0.312 \cdot m o l \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 293.15 \cdot K}{1 \cdot a t m}$ $V=(nRT)/P=(0.312*molxx0.0821*(L*atm)/(K*mol)xx293.15*K)/(1*atm)$

$= 7.5 \cdot L$ $=7.5*L$

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Under conditions of $NTP$ $"NTP"$ , what volume of $S O_{2} (g)$ $SO_2(g)$ results from burning a $10 \cdot g$ $10*g$ mass of elemental sulfur?

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Under conditions of NTP"NTP", what volume of SO2(g)SO_2(g) results from burning a 10⋅g10*g mass of elemental sulfur?

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Under conditions of $NTP$ $"NTP"$ , what volume of $S O_{2} (g)$ $SO_2(g)$ results from burning a $10 \cdot g$ $10*g$ mass of elemental sulfur?