Question

Calculate the volume of nitrogen dioxide produced at 746.0 torr and 20.8°C by the reaction of 6.05 cm³ copper (density = 8.95 g/cm³) with 212.4 mL of concentrated nitric acid if the acid has a density of 1.42 g/cm³ and contains 68.0% HNO₃ by mass).
Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

Answer 1

Here we have two data, so it is a problem of limiting reagent:

We need mass of copper and mass of HNO3 to use:

Copper: 6.05 cm³ copper (density = 8.95 g/cm³)

8.95x6.05= 54.1475 g of Cu

HNO3= 212.4 mL of concentrated nitric acid, has a density of 1.42 g/cm³ and contains 68.0% HNO₃ by mass

First let´s find the mass of the solution= 212.4x1.42= 301.68 g of solution, from this we will find the HNO₃ by multiplying by 68% = 205.09 g of HNO₃

Now wich data we will use?

Form the reaction we can see that for every Cu there has to be at least 4 HNO₃. In mass this will be 63.5 g of Cu (molecular weight) for every 252 g of HNO₃.

Dividing Cu/HNO₃= 63.5/252=0.25 (this is the theorical proportion)

Now dividing our data= 54.1475 / 205.09 = 0.264, higher than the theorical, this means that there is more Cu than HNO₃, we will use the one that is the least.

Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

             252 g    ------------------------------- 2 x 22.4 L

             205.09 -------------------------------- X L

X= 36.46 L of NO₂