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Answer :
The volume of NO₂ gas collected over water at 40.0 °C is calculated to be 69.5 L.
The reaction given is: Cr (s) + 6 HNO₃ (aq) → Cr(NO₃)₃ (aq) + 3 H₂ (l) + 3 NO₂ (g)
First, determine the moles of chromium (Cr): Molar mass of Cr = 52.00 g/mol
Moles of Cr = Mass / Molar mass = 37.9 g / 52.00 g/mol = 0.729 mol Cr
The stoichiometry of the reaction shows that 1 mol of Cr produces 3 mol of NO₂ . Therefore, the moles of NO₂ produced are: Moles of NO₂ = 0.729 mol Cr * 3 mol NO₂ / 1 mol Cr = 2.187 mol NO₂
Next, calculate the partial pressure of NO₂ : Total pressure = 675.0 torr
Vapor pressure of water at 40.0 °C = 55.3 torr
Partial pressure of NO₂ = Total pressure - Vapor pressure of water = 675.0 torr - 55.3 torr = 619.7 torr
Convert the pressure to atm: Pressure in atm = 619.7 torr / 760 torr/atm = 0.815 atm
Using the Ideal Gas Law, PV = nRT, solve for volume (V): R (Ideal Gas Constant) = 0.0821 L·atm/mol·K
Temperature (T) = 40.0 °C = 40.0 + 273.15 = 313.15 K
Volume (V) = nRT / P
V = (2.187 mol) * (0.0821 L·atm/mol·K) * (313.15 K) / (0.815 atm)
V = 69.5 L
The volume of NO₂ gas collected over water at 40.0 °C is 69.5 L.
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The volume of NO₂ gas (in L) collected over water at 40.0 °C when 37.9 g of chromium is added to excess nitric acid is 68.9 L.
What is the volume of NO₂ gas (in L) collected over water at 40.0 °C?
The volume of NO₂ gas (in L) collected over water at 40.0 °C is determined as follows:
Equation of reaction:
Cr (s) + 6 HNO₃ (aq) → Cr(NO₃)₃ (aq) + 3 H₂O (l) + 3 NO₂ (g)
From the equation of reaction, 1 mole of Cr produces 3 moles of NO₂.
Moles NO₂ formed from 37.9 g of Cr will be:
Moles of Cr reacted = 37.9 / 52.0 g = 0.7288 moles
Moles of NO₂ produced = 0.7288 * 3 = 2.1864 moles
The volume of NO₂ produced is calculated from the ideal gas equation:
PV = nRT
Partial pressure of NO₂ = total pressure - pressure of H₂O
Partial pressure of NO₂ = 675 torr - 55.3 torr = 619.7 torr
Partial pressure of NO₂ = 619.7 torr * 1 atm/760 torr = 0.815 atm
Volume, V in L = ?
n = moles = 2.1864 moles
R = gas constant = 0.0821 Latm/Kmol
Temperature in K = 40C + 273 = 313K
V = nRT/P
V = (2.1864) * (0.0821) * (313) / 0.815
Volume = 68.9 L
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