91Ó°ÊÓ

Use the ideal gas law equation to convert the volume of nitrogen gas to moles. The ideal gas law is \[ PV = nRT \] where:-\( P \) is the pressure in atm,- \( V \) is the volume in liters,- \( n \) is the number of moles,- \( R \) is the ideal gas constant, which is 0.0821 \( \text{L·atm}/\text{mol·K} \), and- \( T \) is the temperature in Kelvin.First, convert the pressure from mmHg to atm:\[ 840 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} = 1.1053 \text{ atm} \]Next, convert the temperature from Celsius to Kelvin:\[ 24^{\circ} \text{C} + 273.15 = 297.15 \text{ K} \]Now, use the ideal gas law to solve for moles of \( \text{N}_2 \):\[ n = \frac{PV}{RT} = \frac{1.1053 \text{ atm} \times 2.0 \text{ L}}{0.0821 \text{ L·atm/mol·K} \times 297.15 \text{ K}} = 0.0908 \text{ mol N}_2 \]

Using the stoichiometry of the balanced chemical equation:\[ \text{N}_2(g) + 2 \text{O}_2(g) \rightarrow 2 \text{NO}_2(g) \]For every 1 mole of \( \text{N}_2 \) reacted, 2 moles of \( \text{NO}_2 \) are produced. Therefore, the number of moles of \( \text{NO}_2 \) produced is:\[ 2 \times 0.0908 \text{ mol N}_2 = 0.1816 \text{ mol NO}_2 \]

Use the molar mass of \( \text{NO}_2 \) to convert moles to grams. The molar mass of \( \text{NO}_2 \) is calculated as follows:\[ \text{N} = 14.01 \text{ g/mol} \]\[ \text{O} = 16.00 \text{ g/mol} \times 2 = 32.00 \text{ g/mol} \]\[ \text{NO}_2 = 14.01 + 32.00 = 46.01 \text{ g/mol} \]Now, convert moles of \( \text{NO}_2 \) to grams:\[ 0.1816 \text{ mol NO}_2 \times 46.01 \text{ g/mol} = 8.36 \text{ grams} \]