A container holds an unknown amount of gas with a molar mass of [tex]$98.2 \, \text{g/mol}$[/tex] at a pressure of [tex]$5.5 \, \text{atm}$[/tex]. If the temperature of the gas is [tex]$830.7^\circ \text{C}$[/tex] and the volume is [tex]$4.4 \, \text{liters}$[/tex], what is the mass in grams of the gas in the container? Round your answer to the tenths place.

First, convert the temperature from Celsius to Kelvin:

$$
T = 830.7 + 273.15 = 1103.85~\text{K}.
$$

Next, use the ideal gas law to find the number of moles. The ideal gas law is:

$$
PV = nRT,
$$

which can be rearranged to solve for the number of moles ($n$):

$$
n = \frac{PV}{RT}.
$$

Substitute the known values:

- Pressure, $P = 5.5~\text{atm}$
- Volume, $V = 4.4~\text{L}$
- Ideal Gas Constant, $R = 0.082057~\text{L atm/(mol·K)}$
- Temperature, $T = 1103.85~\text{K}$

Calculate the numerator:

$$
P \times V = 5.5 \times 4.4 = 24.2.
$$

Then the denominator:

$$
R \times T = 0.082057 \times 1103.85 \approx 90.5786.
$$

Now, compute the number of moles:

$$
n = \frac{24.2}{90.5786} \approx 0.26717~\text{mol}.
$$

Next, find the mass of the gas using the molar mass:

$$
\text{mass} = n \times \text{molar mass}.
$$

The molar mass is $98.2~\text{g/mol}$, so:

$$
\text{mass} = 0.26717 \times 98.2 \approx 26.2362~\text{g}.
$$

Finally, rounding the mass to one decimal place gives:

$$
26.2~\text{g}.
$$

Thus, the mass of the gas in the container is $\boxed{26.2~\text{g}}$.