Question

Calculate the volume of nitrogen dioxide produced at 707.4 torr and 22.0°C by the reaction of 8.45 cm3 copper (density = 8.95 g/cm3) with 202.3 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

find moles of HNO₃ used

(202.3 mL) (1.42 g/cm³) = 287.266 grams if conc. nitric acid

(68.0 % of 287.266 grams if conc nitric acid) = 195.341 grams of HNO₃

using molar mass, find moles
(195.341 grams of HNO₃) (1 mol HNO₃ / 63.0130 grams) = 3.1 moles of HNO₃

find moles of Copper used:

(8.45 cm³ copper) (8.95 g/cm3) = 75.6275 grams of copper

using molar mass:
(75.6275 grams of copper) ( 1 mole Cu / 63.546 grams) = 1.19 moles of Copper
find limiting reagent:
by the mole ratios given in the equation
1 Cu(s) + 4 HNO₃(aq) → Cu(NO₃)₂(aq) + 2 NO₂(g) + 2 H₂O(l)
1.19 moles of Copper reacts with 4 times as many moles of HNO₃ - 4.76 moles of HNO₃
the limiting reagent is HNO₃
by the mole ratios given in the equation
1Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)
3.1 moles of HNO₃ produced twice as many moles of NO₂ = 6.2 moles of NO₂


find volume:
PV = nRT
(707.4 Torr) (V) = ( 6.2 moles of NO₂) (62.36 Torr-Litres/mol-K) (295 Kelvin)

V = 161.23 Litres of NO₂