Answer:
To find the volume of nitrogen dioxide gas (\(NO_2\)) formed when 0.129 g of gold (\(Au\)) is dissolved, we first need to determine the number of moles of gold reacted.
1. Calculate the number of moles of gold (\(Au\)):
Given: Mass of gold = 0.129 g
Atomic mass of gold (\(Au\)) = 196.97 g/mol
Number of moles of gold = Mass / Molar mass
\(= 0.129 \, \text{g} / 196.97 \, \text{g/mol}\)
\(= 0.000655 \, \text{mol}\)
2. Use the stoichiometry of the balanced chemical equation to find the moles of \(NO_2\) produced:
From the balanced equation:
1 mole of \(Au\) produces 3 moles of \(NO_2\)
So, \(0.000655 \, \text{mol}\) of \(Au\) will produce \(0.000655 \, \text{mol} \times 3 = 0.001965 \, \text{mol}\) of \(NO_2\).
3. Now, we can use the ideal gas law to find the volume of \(NO_2\) gas produced:
\(PV = nRT\)
Where:
\(P\) = pressure (in atm)
\(V\) = volume (in liters, which we need to find)
\(n\) = number of moles of gas
\(R\) = ideal gas constant (\(0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K}\))
\(T\) = temperature (in Kelvin)
Given: Pressure (\(P\)) = 1.000 atm
Temperature (\(T\)) = 294.65 K
Number of moles of \(NO_2\) (\(n\)) = 0.001965 mol
Ideal gas constant (\(R\)) = 0.0821 atm·L/(mol·K)
Substituting the values into the equation:
\(V = \frac{nRT}{P}\)
\(V = \frac{0.001965 \, \text{mol} \times 0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K} \times 294.65 \, \text{K}}{1.000 \, \text{atm}}\)
\(V ≈ 0.0475 \, \text{L}\)
So, the volume of nitrogen dioxide gas formed when 0.129 g of gold is dissolved in excess aqua regia is approximately 0.0475 liters.
