What volume of nitrogen dioxide (in L) is produced from the complete reaction of 16.87 of lead (ii) nitrate at STP

figure the moles

check the balanced equation to see how many moles of NO2 are produced.

Recall that 1 mole fills 22.4L at STP

Thank you so much!

To find the volume of nitrogen dioxide produced, we need to first calculate the moles of lead (II) nitrate and then use the stoichiometry of the balanced equation to determine the moles of nitrogen dioxide produced. Finally, we can convert the moles of nitrogen dioxide to liters at STP.

Let's start by calculating the moles of lead (II) nitrate.

1. Calculate the molar mass of lead (II) nitrate (Pb(NO3)2):
Pb: 207.2 g/mol
N: 14.0 g/mol
O: 16.0 g/mol (3 oxygen atoms in one nitrate group)

Molar mass of Pb(NO3)2 = (207.2 g/mol) + (2 * (14.0 g/mol + 16.0 g/mol))
= 331.2 g/mol

2. Convert the given mass of lead (II) nitrate to moles using its molar mass:
Moles of Pb(NO3)2 = mass / molar mass
= 16.87 g / 331.2 g/mol
≈ 0.0509 mol

Next, let's determine the stoichiometry of the balanced equation between lead (II) nitrate and nitrogen dioxide:

3 Pb(NO3)2 -> 3 PbO + 6 NO2 + O2

From the balanced equation, we can see that for every 3 moles of Pb(NO3)2, 6 moles of NO2 are produced.

3. Calculate the moles of nitrogen dioxide produced:
Moles of NO2 = (moles of Pb(NO3)2) * (6 moles of NO2 / 3 moles of Pb(NO3)2)
= 0.0509 mol * (6 / 3)
= 0.1018 mol

Finally, let's convert the moles of nitrogen dioxide to liters at STP:

4. Use the ideal gas law equation:
PV = nRT

At STP, the temperature (T) is 273.15 K, and the pressure (P) is 1 atm.

Rearranging the equation:
V = (n * R * T) / P

Substituting the values:
V = (0.1018 mol * 0.0821 L*atm/mol*K * 273.15 K) / 1 atm
= 2.2782 L

Therefore, the volume of nitrogen dioxide produced from the complete reaction of 16.87 g of lead (II) nitrate at STP is approximately 2.2782 Liters.

To find the volume of nitrogen dioxide (NO2) produced from the complete reaction of lead (II) nitrate (Pb(NO3)2), we need to use the balanced chemical equation and the ideal gas law.

1. Write the balanced chemical equation:
2Pb(NO3)2 → 4NO2 + O2 + 2Pb

From the balanced equation, we can see that for every 2 moles of Pb(NO3)2, we will get 4 moles of NO2.

2. Convert the mass of lead (II) nitrate to moles:
Given that the mass of Pb(NO3)2 is 16.87 g, we can calculate the number of moles using its molar mass. The molar mass of Pb(NO3)2 can be calculated as follows:

Pb: (1 atom) 207.2 g/mol
N: (2 atoms) 2 × 14.0 g/mol = 28.0 g/mol
O: (6 atoms) 6 × 16.0 g/mol = 96.0 g/mol

Molar mass of Pb(NO3)2 = 207.2 + 28.0 + 96.0 = 331.2 g/mol

Moles of Pb(NO3)2 = mass / molar mass
Moles of Pb(NO3)2 = 16.87 g / 331.2 g/mol

3. Calculate moles of NO2 produced:
According to the balanced equation, for every 2 moles of Pb(NO3)2, we get 4 moles of NO2. Therefore, the moles of NO2 produced can be calculated using the mole ratio:

Moles of NO2 = Moles of Pb(NO3)2 × (4 moles of NO2 / 2 moles of Pb(NO3)2)

4. Convert moles of NO2 to volume at STP:
To determine the volume of NO2 at STP (standard temperature and pressure), we can use the ideal gas law:

PV = nRT

Where:
P = pressure (at STP, 1 atm)
V = volume of gas
n = number of moles
R = ideal gas constant (0.0821 L.atm/(mol.K))
T = temperature (at STP, 273.15 K)

Rearranging the ideal gas law equation, we can solve for volume:

V = nRT / P

Considering that at STP, the pressure (P) is 1 atm, and the temperature (T) is 273.15 K, we can substitute these values into the equation:

V = (moles of NO2) × (0.0821 L.atm/(mol.K)) × (273.15 K) / 1 atm

Now you can substitute the value of moles of NO2 to calculate the volume.