Question

Calculate the volume of nitrogen dioxide produced at 789.4 torr and 26.9°C by the reaction of 8.55 cm3 copper (density = 8.95 g/cm3) with 221.3 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

First we have to find moles of HNO3 used

GGiven volume of HNO3=221.3 mL, density=1.42 g/mL, and molar mass of HNO3=63.01 g/mol

(221.3 mL) (1.42 g/cm3) = 314.246 grams of con nitric acid

But here we have 68% HNO3, (0.68x314.246 grams of con nitric acid) = 213.687 grams of HNO3

using molar mass, find moles
(213.687 grams of HNO3) (1 mol HNO3 / 63.0130 grams) = 3.391 moles of HNO3
Now we will find moles of Copper used:
(8.55 cm3 copper) (8.95 g/cm3) = 76.5225 grams of copper
using molar mass:
(76.5225 grams of copper) ( 1 mole Cu / 63.546 grams) = 1.204 moles of Copper

Now find limiting reagent:
by the mole ratios given in the equation
1 mole Cu(s) + 4 moles HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
1.204 moles of Copper reacts with 4 times as many moles of HNO3 - 4.8168 moles of HNO3
they added an excess of Cu (s), when they used 3.391 moles of HNO3
the limiting reagent is HNO3

Then from the equation the mole ratios
1 mole Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2 moles NO2(g) + 2H2O(l)
4 moles of HNO3 gives 2 moles of NO2, moles of NO2 = 1/2(3.391moles of HNO3)=1.6955 moles of NO2.
Now find volume:
PV = nRT
(789.4 Torr) (V) = ( 1.6955 moles of NO2) (62.36 Torr-Litres/mol-K) (299.9 Kelvin)

V = 40.1682 Litres of NO2

So volume V=40 L.