Answer:
The volume of nitrogen dioxide formed is 14.7 L
Step-by-step explanation:
Step 1: Data given
Pressure = 708 torr
Temperature = 28.2 °C
volume of copper = 1.95 cm³ of copper
Density of copper = 8.95 g/cm
Volume of 2.53 M nitric acid = 1.50 L
Step 2: The balanced equation
Cu(s) + 4 HNO3 (aq) → Cu(NO3)2(aq) + 2NO2(g) + 2 H2O
Step 3: Calculate the moles of HNO3
Moles HNO3 = Molarity * volume
Moles HNO3 = 2.53 M * 1.50 L
Moles HNO3 = 3.795 moles
Step 4: Calculate mass of copper
Mass = density * volume
Mass coper = 8.95 g/cm³ * 1.95 cm³
Mass of copper = 17.4525 grams copper
Step 5: Calculate moles Cu
Moles Cu = mass of Cu / molar mass Cu
Moles Cu = 17.4525g / 63.546 g/mol
Moles Cu = 0.27464 moles
Step 6: Calculate limiting reactant
For 1 mole of copper, we need 4 moles of HNO3 to produce 2 moles of NO2
Cu is the limiting reactant. It will completely be consumed (0.27464 moles).
HNO3 is in excess. There will 4*0.27464 = 1.09856 moles be consumed.
There will remain 3.795 - 1.09856 = 2.69644 moles
Step 7: Calculate moles of NO2
For 1 mole of copper, we need 4 moles of HNO3 to produce 2 moles of NO2
For 1 moles 0.27464 moles of Cu, there will be produced 2* 0.27464 = 0.54928 moles of NO2
Step 8: Calculate volume of NO2
p*V = n*R*T
⇒ p = the pressure = 705 torr = 705/760 = 0.927632 atm
⇒ V = the volume of NO2 = TO BE DETERMINED
⇒ n = the moles of NO2 = 0.54928 moles
⇒ R = the gas constant = 0.08206 L*atm/K*mol
⇒ T = the temperature = 28.2 °C = 301.35 Kelvin
V = (nRT)/p
V = (0.54928 *0.08206 * 301.35)/0.927632
V = 14.7 L
The volume of nitrogen dioxide formed is 14.7 L