Consider the reaction for the production of NO2 from NO:
2NO(g) + O2(g) → 2 NO2(g)
(a) If 84.8 L of O2(g), measured at 35 °C and 632 mm Hg, is allowed to react with 158.2 g of NO, find the limiting reagent.
Volume of O2 = 84.8 L
Pressure = 632 mm Hg
Temperature = 35 oc
Mass of NO = 158.2 g
Here we will have to find out the limiting reagent
The given equation for the reaction is ,
2NO(g) + O2(g) → 2NO2(g)
The way to solve this is to find the moles of O2 and NO2 that we have, to find moles O2: PV = nRT so n = PV/RT,
where T = 35oC + 273 = 308 K and R = 62.36 L-mmHg/mol-K
n = (632 mmHg)(84.8L)/(62.36)(308K)...