What volume of dry NO at 25°C and 1.00 atm can be prepared by reacting 6.35 g of Cu with 25.0 mL of 6.0 M nitric acid, HNO3?
3 Cu(s) + 8 HNO3(aq) --> 3 Cu(NO3)2(aq) + 2 NO(g)+ 4 H2O(l)
To find the volume of dry NO that can be prepared, we need to use stoichiometry to determine the moles of NO produced and then convert that to volume using the ideal gas law.
First, let's calculate the moles of Cu(NO3)2 produced. Start by finding the moles of Cu reacted:
Molar mass of Cu = 63.55 g/mol
Moles of Cu = mass / molar mass
Moles of Cu = 6.35 g / 63.55 g/mol
Next, use the balanced chemical equation to determine the moles of Cu(NO3)2 produced:
3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)
From the equation, we can see that 3 moles of Cu react to produce 2 moles of NO. Therefore, the moles of Cu(NO3)2 is equal to the moles of Cu multiplied by a ratio of 2 moles of NO per 3 moles of Cu:
Moles of Cu(NO3)2 = (Moles of Cu) × (2 moles of NO / 3 moles of Cu)
Now, we can calculate the moles of NO produced:
Moles of NO = Moles of Cu(NO3)2 × (2 moles of NO / 3 moles of Cu(NO3)2)
Next, we need to convert the moles of NO to volume using the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature (in Kelvin)
Since we want the volume at 25°C (298 K) and 1.00 atm, we can rearrange the ideal gas law to solve for V:
V = (nRT) / P
Now, we can substitute the values into the equation:
V = (Moles of NO) × (R) × (T) / P
Calculate the volume of NO:
V = (Moles of NO) × (0.0821 L·atm/mol·K) × (298 K) / 1.00 atm
This will give you the volume of NO in liters.
This is a limiting reagent (LR) problem; you know that because amounts are given for BOTH reactants.
mols Cu = grams/atomic mass = ?
mols HNO3 = M x L = ?
Using the coefficients in the balanced equation, convert mols Cu to mols NO.
Do the same and convert mols HNO to mols NO.
You likely will obtain different values for mols NO which means one of the values is wrong; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that is the LR.
Using the smaller value for mols NO, to volume using PV = nRT