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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 moles of $\ce{Xe}$ using the ideal gas law:

$$n = \frac{PV}{RT} = \frac{\pu{0.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.18 mol} \ \ce{Xe}$$

Doing the same for fluorine gives:

$$ n = \frac{\pu{1.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.54 mol}\ \ce{F2}$$

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  :

$$ \pu{0.18 mol} \ \ce{XeF4} \cdot \pu{207.1 g mol-1} \ \ce{XeF4} = \pu{37.278 g}\ \ \ce{XeF4}$$

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?

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?

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?

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 moles of $\ce{Xe}$ using the ideal gas law:

$$n = \frac{PV}{RT} = \frac{\pu{0.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.18 mol} \ \ce{Xe}$$

Doing the same for fluorine gives:

$$ n = \frac{\pu{1.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.54 mol}\ \ce{F2}$$

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 :

$$ \pu{0.18 mol} \ \ce{XeF4} \cdot \pu{207.1 g mol-1} \ \ce{XeF4} = \pu{37.278 g}\ \ \ce{XeF4}$$

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?