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original question: if we burn 390g of poly styrene (STP), how much gas produces?

me: 1008L

my teacher: 672L

he says h2o is not in gas form!(because of STP) who is wrong?

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    $\begingroup$ Neither of you is correct. The problem is indeterminate assuming that after the reaction the system was brought back to STP because the initial volume of the gas was not specified. If say initially a million liters of oxygen was used then all the water would be present as the vapor. However if a stoichiometric amount of oxygen was used then most, but not all, of the water would be in liquid form $\endgroup$ – MaxW May 12 at 20:04
  • $\begingroup$ You can't burn polystyrene at STP. You didn't state the entire problem so the final conditions are a bit of a mystery too. But it would seem that the system (products after combustion) was brought back to STP after the reaction. $\endgroup$ – MaxW May 12 at 21:25
  • $\begingroup$ @MaxW why we can't burn? $\endgroup$ – aStudent May 12 at 21:30
  • $\begingroup$ You can't burn polystyrene at STP. i.e. constantly at $\pu{O ^\circ C}$. When you burn polystyrene it gets hot. So you then have to cool the products down. $\endgroup$ – MaxW May 12 at 21:40
  • $\begingroup$ @MaxW so we couldn't burn things while we are near sea and temp is O°C? $\endgroup$ – aStudent May 12 at 21:50
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Neither of you is correct. The problem is indeterminate assuming that after the reaction the system was brought back to STP because the initial volume of the gas was not specified. If say initially a million liters of oxygen was used then all the water would be present as the vapor. However if a stoichiometric amount of oxygen was used then most, but not all, of the water would be in liquid form

$$\ce{\pu{M.W. }\ C8H8 = \pu{104.15 g/mol}}\tag{1}$$

$$\ce{\pu{mol }\ C8H8 =\dfrac{\pu{390 g}}{\pu{104.15 g/mol}}} = \pu{3.7446 mol} \tag{2}$$

$$\ce{C8H8 ->[O2] 8CO2 + 4H2O}\tag{3}$$

1.000 mole of a gas at STP ( $\pu{0 °C}$ and $\pu{100 kPa}$) has a volume of $\pu{22.711 L/mol}$.

Let's suppose that a huge excess of oxygen gas was used for the combustion. After the combustion the system is brought back to STP, and all the water is present as vapor. So the upper limit of the gaseous products is:

$$V_\mathrm{gas, UL} =3.7446 \times 12\times 22.711 = \pu{1020.52 L} \ce{->[round to 3] \pu{1.20\times 10^3 L}}\tag{4}$$

Now suppose that a stoichometric amount of oxygen gas was used for the combustion. After the combustion the system is brought back to STP, and some of the water is present as vapor, but most is present as the liquid. The vapor pressure of water at $\pu{0 °C}$ is $\pu{0.6113 kPa}$. So the lower limit of the gaseous products is:

\begin{align} V_\mathrm{gas, LL} &= (3.7446 \times 8\times 22.711) \times \left(\dfrac{\pu{100 kPa}}{\pu{100 kPa} - \pu{0.6113 kPa}}\right) \\ &= (680.35)\times (1.00615) = \pu{684.53 L} \ce{->[round to 3]} \pu{685 L}\\ \end{align}

Old definition of STP

Before 1983 the definition of STP was $\pu{0 ^\circ C}$ and $\pu{101.325 kPa}$ for which 1.000 mole of a gas has a volume of 22.414 L/mol. Using the old value:

$$V_\mathrm{gas, UL} = \dfrac{22.414}{22.711}\times \pu{1020.52 L}= \pu{1007 L}$$

$$V_\mathrm{gas, LL} = \dfrac{22.414}{22.711}\times \pu{684.53 L}= \pu{676 L}$$


The OP finally made me realize that the above answer for $V_\mathrm{gas, LL}$ was not correct for the old STP.

The problem lies with the fact that the pressure correction for the vapor pressure of water involves subtraction and not just multiplication. As the OP points out:

$$\dfrac{\pu{100 kPa}}{\pu{100 kPa} - \pu{0.6113 kPa}} \ne \dfrac{\pu{101.325 kPa}}{\pu{101.325 kPa} - \pu{0.6113 kPa}}$$

So my calculation yielded

$$V_\mathrm{gas, LL} = \dfrac{22.414}{22.711}\times \pu{684.53 L}= 675.58 \ce{->[Round(3)]}\pu{676 L}$$

But the correct calculation is:

\begin{align} V_\mathrm{gas, LL} &= (3.7446 \times 8\times 22.414) \times \left(\dfrac{\pu{101.325 kPa}}{\pu{101.325 kPa} - \pu{0.6113 kPa}}\right) \\ &= (671.45)\times (1.00607) = \pu{675.52 L} \ce{->[round to 3]} \pu{676 L}\\ \end{align}

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  • $\begingroup$ could you explain about 100kPa/(100kPa - 0.6113kPa) ? $\endgroup$ – aStudent May 14 at 18:58
  • $\begingroup$ @aStudent - The current specification for STP defines the pressure as 100 kPa. If a stoichiometric amount of oxygen was used then the gas would contain $\ce{CO2}$ and $\ce{H2O}$. at $0 ^\circ \mathrm{C}$ the vapor pressure of water alone is 0.6113kPa. The vapor pressure of $\ce{CO2}$ alone would be $\pu{100kPa - 0.6113kPa}$. $\endgroup$ – MaxW May 14 at 19:19
  • $\begingroup$ oh excuse me! i made a small mistake. i will delete my comment and i will tell you after checking. $\endgroup$ – aStudent May 15 at 21:13
  • $\begingroup$ Let's get back to basics. Was the volume of the products measured at STP, ($\pu{0 ^\circ C}$ and $\ce{100 kPa}$) or not? If not what T and P? $\endgroup$ – MaxW May 15 at 21:14
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    $\begingroup$ we should do three corrections : (3.7446×8× 22.414 )×( 100 kPa /( 100 kPa −0.6113 kPa)) .(we should also change pressure of CO2 and total gases) i.e. : ( 100 kPa /( 100 kPa −0.6113 kPa)) does not exactly equal to ( 101.325 kPa /( 101.325 kPa −0.6113 kPa)) $\endgroup$ – aStudent May 15 at 23:16

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