# 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?

## Expert Answers 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)`

At STP,

`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`

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