Understanding gas stoichiometry for the reaction of xenon and fluorine

Question

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?

Answer 1

Yes, this is a rounding 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.

So, in your final equation:

$$\pu{0.181072885 mol} \times \pu{207.1 g//mol} = \pu{37.5 g}$$

I find using the full value of an intermediate step usually eliminates these types of errors in the final calculation.