# Understanding gas stoichiometry for the reaction of xenon and fluorine

A $$\pu{20.0 L}$$ nickel container was charged with $$\pu{0.500 atm}$$ of xenon gas and $$\pu{1.50 atm}$$ of fluorine gas at $$\pu{400 ^{\circ}C}$$. 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 amount of substance of $$\ce{Xe}$$ using the ideal gas law: $$n(\ce{Xe}) = \frac{pV}{RT} = \frac{\pu{0.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.18 mol}$$

Doing the same for fluorine gives: $$n(\ce{F2}) = \frac{\pu{1.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.54 mol}$$

Then, for every mole of $$\ce{Xe}$$ we need two moles of $$\ce{F2}$$. Since we have more than $$2 \cdot \pu{0.18 mol} = \pu{0.36 mol}$$ 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: \begin{align} n(\ce{XeF4}) M(\ce{XeF4}) &= m(\ce{XeF4})\\ \pu{0.18 mol} \times \pu{207.1 g mol-1} &= \pu{37.278 g} \end{align}

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

Where you calculate: $$n = \frac{pV}{RT} = \frac{\pu{0.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.18 mol}$$
The actual unrounded answer is $$\pu{0.181072885 mol}$$ - a little bit higher than the rounded value.
$$\pu{0.181072885 mol} \times \pu{207.1 g//mol} = \pu{37.5 g}$$