An STP volume of 564 liters of oxygen is produced in the reaction 2 N₂O₅(g)$\rightarrow$ 4 NO₂(g)+ O₂(g).What volume of nitrogen dioxide is also produced if it is measured at 25°C and 1.00 bar?

Question

Quizplus · Accepted Answer

The ideal gas law can be used to solve this problem. PV = nRTwhere P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.At STP, the temperature is 273.15 K and the pressure is 1 bar. The volume of 564 liters of oxygen is produced in the reaction.2 N₂O₅(g) → 4 NO₂(g) + O₂(g)This means that for every 2 moles of N₂O₅ that react, 4 moles of NO₂ are produced.So, the number of moles of NO₂ produced is:n(NO₂) = 4/2 * n(N₂O₅)The number of moles of N₂O₅ that react is:n(N₂O₅) = PV/RT = (1 bar * 564 L)/(0.0821 Latm/(molK) * 273.15 K) = 22.4 molSo, the number of moles of NO₂ produced is:n(NO₂) = 4/2 * 22.4 mol = 44.8 molThe volume of NO₂ produced at 25°C and 1.00 bar is:V(NO₂) = nRT/P = (44.8 mol * 0.0821 Latm/(molK) * 298.15 K)/1 bar = 1030 LSo, the correct answer is A) 2.46 × 10³ L.

An STP Volume of 564 Liters of Oxygen Is Produced →\rightarrow→

An STP Volume of 564 Liters of Oxygen Is Produced $\rightarrow$