i have the Mr of lead nitrate 331 and i thermally decompose it giving nitrogen dioxide and oxygen. i want to collect exactly 200cm3 of the gases evolved in a syringe. what mass of lead nitrate should i take ?
Here is a worked example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
can you do it for me if possible drbob222 ?
I will be glad to help you through it if you have trouble. Post your work and explain what you don't understand.
To determine the mass of lead nitrate needed to produce exactly 200 cm3 of the gases evolved, we need to calculate the molar ratio between lead nitrate, nitrogen dioxide (NO2), and oxygen (O2).
Let's start by writing the balanced chemical equation for the thermal decomposition of lead nitrate:
2 Pb(NO3)2 -> 2 PbO + 4 NO2 + O2
From the equation, we can see that 2 moles of lead nitrate (Pb(NO3)2) produce 4 moles of nitrogen dioxide (NO2) and 1 mole of oxygen (O2).
1 mole of any gas at standard temperature and pressure (STP) occupies approximately 22.4 liters or 22,400 cm3.
Since we want to collect exactly 200 cm3 of the evolved gases, we need to determine the volume of each gas produced.
For nitrogen dioxide (NO2), we need 200 cm3. As the molar ratio is 4:1, the volume of nitrogen dioxide at STP would be:
200 cm3 * (1 mole NO2 / 4 moles NO2) * (22,400 cm3 / 1 mole NO2) = 11,200 cm3
For oxygen (O2), the molar ratio is 1:1, so we also need 200 cm3 of oxygen.
Now, to calculate the moles of each gas, we can use the ideal gas law:
PV = nRT
Where:
P = pressure (at STP, this is approximately 1 atmosphere)
V = volume in liters (converted from cm3 by dividing by 1000)
n = moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin (at STP, this is 273.15 K)
Using the ideal gas law, we can solve for the moles of each gas:
nNO2 = (P * VNO2) / (R * T)
nO2 = (P * VO2) / (R * T)
Since the pressures and temperatures are the same, we can ignore those variables.
nNO2 = VNO2 / 22.4
nO2 = VO2 / 22.4
Substituting the volumes we need in cm3:
nNO2 = 11,200 / 22.4 = 500 moles
nO2 = 200 / 22.4 = 8.93 moles
From the balanced equation, we know that 2 moles of lead nitrate produce 4 moles of nitrogen dioxide. Therefore, to produce 500 moles of nitrogen dioxide, we need:
(moles of Pb(NO3)2) = (moles of NO2) * (moles of Pb(NO3)2 / moles of NO2) = (500/4) = 125 moles
Finally, to calculate the mass of lead nitrate needed, we use the molar mass of lead nitrate (Pb(NO3)2):
mass = moles * molar mass
mass = 125 moles * molar mass (Pb(NO3)2)
The molar mass of lead nitrate (Pb(NO3)2) can be calculated by adding up the atomic masses of its elements:
molar mass (Pb(NO3)2) = (molar mass Pb) + 2 * (molar mass N) + 6 * (molar mass O)
molar mass (Pb(NO3)2) = (207.2 g/mol) + 2 * (14.0 g/mol) + 6 * (16.0 g/mol)
molar mass (Pb(NO3)2) = 331.2 g/mol
Substituting this back into the mass equation:
mass = 125 moles * 331.2 g/mol
Therefore, you should take approximately 41,400 grams (or 41.4 kilograms) of lead nitrate to produce exactly 200 cm3 of the gases evolved.