Hydrogen chloride gas is shipped in a container under 5 ,100 mm Hg of pressure that occupies 201 liters at 29^circ C How many liters of gas would be produced at STP? 8.3times 10^-3L 150 L 1201 17times 10^-5L

To determine the volume of hydrogen chloride gas at standard temperature and pressure (STP), we can use the combined gas law, which is given by:<br /><br />\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]<br /><br />where:<br />- \( P_1 \) is the initial pressure,<br />- \( V_1 \) is the initial volume,<br />- \( T_1 \) is the initial temperature in Kelvin,<br />- \( P_2 \) is the final pressure (STP),<br />- \( V_2 \) is the final volume (at STP),<br />- \( T_2 \) is the final temperature in Kelvin (STP).<br /><br />Given data:<br />- \( P_1 = 5100 \) mm Hg<br />- \( V_1 = 201 \) liters<br />- \( T_1 = 29^\circ C = 29 + 273.15 = 302.15 \) K<br />- \( P_2 = 760 \) mm Hg (standard pressure)<br />- \( T_2 = 0^\circ C = 273.15 \) K (standard temperature)<br /><br />We need to find \( V_2 \).<br /><br />Rearranging the combined gas law to solve for \( V_2 \):<br /><br />\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]<br /><br />Substituting the given values:<br /><br />\[ V_2 = \frac{5100 \times 201 \times 273.15}{760 \times 302.15} \]<br /><br />Calculating the numerator and the denominator separately:<br /><br />Numerator: \( 5100 \times 201 \times 273.15 = 280,497,015 \)<br /><br />Denominator: \( 760 \times 302.15 = 229,634 \)<br /><br />Now, divide the numerator by the denominator:<br /><br />\[ V_2 = \frac{280,497,015}{229,634} \approx 1221.5 \text{ liters} \]<br /><br />Therefore, the volume of hydrogen chloride gas at STP would be approximately 1221.5 liters.<br /><br />Among the given options, the closest value is:<br /><br />1201<br /><br />So, the correct answer is:<br /><br />1201