Ethene is converted to ethane by the reaction $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)\stackrel{\text { Catalyst }}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}(g)$$ \(\mathrm{C}_{2} \mathrm{H}_{4}\) flows into a catalytic reactor at 25.0 \(\mathrm{atm}\) and \(300.^{\circ} \mathrm{C}\) with a flow rate of $1000 . \mathrm{L} / \mathrm{min}\( . Hydrogen at 25.0 \)\mathrm{atm}\( and \)300 .^{\circ} \mathrm{C}\( flows into the reactor at a flow rate of \)1500 . \mathrm{L} / \mathrm{min}\( . If 15.0 \)\mathrm{kg}\( \)\mathrm{C}_{2} \mathrm{H}_{6}$ is collected per minute, what is the percent yield of the reaction?

First, we need to find the number of moles flowed in for both ethene (C2H4) and hydrogen (H2). Using the ideal gas law equation: \(PV = nRT\), we can find the number of moles, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, convert the given temperature (300°C) to Kelvin: \(T_{K} = 300 + 273.15 = 573.15 \: K\) Now, we calculate the moles of ethene: Formula of ethene is \(C_{2}H_{4}\), and its molar mass is 28.05 g/mol. \(P_{C2H4} = 25\: atm\) \(V_{C2H4} = 1000\: L/min\) \(T_{C2H4} = 573.15 \: K\) \(R = 0.0821\: L.atm/mol.K\) \(n_{C2H4} = \frac{P_{C2H4}V_{C2H4}}{RT_{C2H4}} = \frac{25 \times 1000}{0.0821 \times 573.15} = 684.74\: moles/min\) Similarly, we calculate the moles of hydrogen: Formula of hydrogen is \(H_{2}\), and its molar mass is 2.02 g/mol. \(P_{H2} = 25\: atm\) \(V_{H2} = 1500\: L/min\) \(T_{H2} = 573.15 \: K\) \(n_{H2} = \frac{P_{H2}V_{H2}}{RT_{H2}} = \frac{25 \times 1500}{0.0821 \times 573.15} = 1027.11\: moles/min\)

As per the balanced equation of given reaction: \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)\longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\) Thus, the molar ratio of ethene to hydrogen in the reaction is 1:1. From Step 1, we know: \(n_{C2H4} = 684.74 \: moles/min\) \(n_{H2} = 1027.11 \: moles/min\) Since ethene is the limiting reactant, the reaction proceeds until all of the ethene is used up. After this, there will be no ethene remaining, and the unreacted hydrogen will be: \(n_{H2} - n_{C2H4} = 1027.11 - 684.74 = 342.37\: moles/min\)

When all ethene is used up, the theoretical moles of ethane produced are equal to the moles of ethene. \(n_{C2H6} = n_{C2H4} = 684.74\: moles/min\) Now, let's find the theoretical mass of ethane produced. Formula of ethane is \(C_{2}H_{6}\), and its molar mass is 30.07 g/mol. \(Mass_{C2H6} = n_{C2H6} * Molar\:mass_{C2H6} = 684.74 * 30.07 = 20584.20 g/min\)

Now, we have the theoretical mass of ethane produced, and we are given that the actual mass of ethane collected per minute is 15.0 kg = 15000 g. Percent yield of reaction is calculated as: \(\%\:yield = \frac{Actual\:mass\:of\:product}{Theoretical\:mass\:of\:product} * 100\) \(\%\:yield = \frac{15000}{20584.20} * 100 = 72.89\%\) The percent yield of the reaction is approximately 72.89%.