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What volume of $O_2(g)$ at 350. $"^o$C and a pressure of 5.25 atm is needed to completely convert 10.0 g of sulfur to sulfur trioxide?

Md Ariful Islam

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Md Ariful Islam

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$"Moles of sulfur"$ $=$ $(10.0*g)/(32.06*g*mol^-1)=0.312*mol$.

Given the , we require $3/2xx0.312*mol$ of dioxygen gas, i.e. $0.468*mol$.

And now we solve for volume in the Ideal Gas equation:

$V=(nRT)/P$ $=$ $(0.468*cancel(mol)xx0.0821*L*cancel(atm)*cancel(K^-1)*cancel(mol^-1)xx623*cancel(K))/(5.25*cancel(atm))$ $~=$ $5*L$.

Note (i) that we converted $"degrees Celsius"$ to $"degrees Kelvin"$, and (ii) that we had to use an appropriate $"Gas constant, R"$. $"R"$ with various units would be supplied as supplementary material in any exam. The fact that our equation gave units of volume, when a volume was sought in the problem, can help to convince us that we got the problem right (for once!!).

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