# Understanding gas stoichiometry for the reaction of xenon and fluorine

A $20.0$-L nickel container was charged with $0.500$ atm of xenon gas and $1.50$ atm of fluorine gas at $400$ $^{\circ}$C. The xenon and fluorine react to form xenon tetrafluoride. What mass of xenon tetrafluoride can be produced assuming $100$ % yield?

Attempt at solution: From the description, the reaction can be written as $$\ce{Xe + 2F2 -> XeF4}$$ I first calculated the moles of $\ce{Xe}$ using the ideal gas law: $$n = \frac{PV}{RT} = \frac{0.5 \ \text{atm} \cdot 20.0 \ \text{L}}{0.08206 \cdot 673\, \text{K}} = 0.18 \ \text{mol} \ \ce{Xe}$$

Doing the same for fluorine gives:

$$n = \frac{1.5 \ \text{atm} \cdot 20.0 \ \text{L}}{0.08206 \cdot 673\, \text{K}} = 0.54 \ \text{mol} \ \ce{F2}$$

Then, for every mole of $\ce{Xe}$ we need two moles of $\ce{F2}$. Since we have more than $2 \cdot 0.18 \ \text{mol} = 0.36 \ \text{mol}$ of fluorine, xenon is the limiting reactant. The molar mass of $\ce{XeF4}$ is $207.1\,\mathrm{g/mol}$.

So now I would just do :

$$0.18 \ \text{mol} \ \ce{XeF4} \cdot \frac{207.1\, \mathrm{g} \ \ce{XeF4}}{\text{mol}}$$ which gives me $37.278\,\mathrm{g}$ of $\ce{XeF4}$.

However, the answer at the back of my textbook says it should be $37.5\,\mathrm{g}$. So did I make a mistake somewhere or is this just a roundoff-error?

Yes, this is a rounding error.

Where you calculate:

$$n = \frac{PV}{RT} = \frac{0.5 \ \text{atm} \cdot 20.0 \ \text{L}}{0.08206 \cdot 673 \text{K}} = 0.18$$

The actual unrounded answer is $0.181072885$ - a little bit higher than the rounded value.

$0.181072885 \times 207.1 = 37.5g$