Cu + HNO3 -> Cu(NO3)2 + NO2+ H2O How many liters of NO2 gas at STP will be formed by the reaction of 75.5 g of copper with 55.0 g of nitric acid?
First we need to determine the limiting reagent.
`Cu + 4 HNO_3 -> Cu(NO_3)_2 + 2 NO_2 + 2 H_2O`
`75.5 grams Cu * (1 mol e Cu)/(63.546 grams Cu) * (2 mol es NO_2)/(1 mol e Cu)`
`= 2.376 mol es NO_2`
`55.0 grams HNO_3 * (1 mol e HNO_3)/(63.01 grams HNO_3) * (2 mol es NO_2)/(4 mol es HNO_3)`
`= 0.4364 mol es NO_2`
Looking at the results, we can see that the limiting reagent is the nitric acid and therefore we will use the data derived from the nitric acid.
`n = 0.4364 mol es NO_2`
`R = 0.08206 (atm-L)/(mol-K)`
`T = 273.15 K`
`P = 1 atm`
Using the ideal gas law we can determine the volume that the NO2 occupies.
`PV = nRT`
`V = (nRT)/(P)`
`V = (0.4364 *0.08206* 273.15)/(1)`
`V = 9.78 L`