How many grams of phosgenite can be obtained from 14.0 g of PbO and 14.0 g of NaCl in the presence of excess water and carbon dioxide?
This is a limiting reagent (LR) problem. We know that because amounts are given for more than one reactant.
Write the equation and balance it.
Convert grams PbO to mols. mols = grams/molar mass
Do the same for NaCl.
Using the coefficients in the balanced equation, convert mols PbO to mols phosgenite.
Do the same for mols NaCl to mols phosgenite.
It is likely that these two values will not agree which means one of them is wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent responsible for the smaller number is the LR.
Now use the smaller value for mols and convert to grams. g = mols x molar mass.
To calculate the amount of phosgenite (Pb2Cl2CO3) that can be obtained, we need to find the limiting reagent between PbO and NaCl. The limiting reagent is the one that gets completely consumed and determines the maximum amount of product that can be formed.
First, let's find the moles of each reactant:
Molar mass of PbO:
Pb = 207.2 g/mol
O = 16.0 g/mol
Total molar mass = 207.2 + 16.0 = 223.2 g/mol
Moles of PbO = Mass of PbO / Molar mass of PbO
Moles of PbO = 14.0 g / 223.2 g/mol
Molar mass of NaCl:
Na = 22.99 g/mol
Cl = 35.45 g/mol
Total molar mass = 22.99 + 35.45 = 58.44 g/mol
Moles of NaCl = Mass of NaCl / Molar mass of NaCl
Moles of NaCl = 14.0 g / 58.44 g/mol
Next, we need to determine the stoichiometric ratio between PbO and Pb2Cl2CO3:
The balanced chemical equation is:
2PbO + 2NaCl + H2O + CO2 -> 2Pb2Cl2CO3 + 2NaOH
From the stoichiometry, we see that 2 moles of PbO react with 2 moles of NaCl to produce 2 moles of Pb2Cl2CO3.
Now, let's calculate the moles of Pb2Cl2CO3 that can be formed:
Moles of Pb2Cl2CO3 = Moles of limiting reagent (PbO) / Stoichiometric ratio
Moles of Pb2Cl2CO3 = (14.0 g / 223.2 g/mol) / 2
Finally, we can calculate the mass of Pb2Cl2CO3 in grams:
Mass of Pb2Cl2CO3 = Moles of Pb2Cl2CO3 x Molar mass of Pb2Cl2CO3
Mass of Pb2Cl2CO3 = [(14.0 g / 223.2 g/mol) / 2] x (303.08 g/mol)
Simplifying the equation gives us:
Mass of Pb2Cl2CO3 = 0.0626 g (rounded to four decimal places)
Therefore, approximately 0.0626 grams of phosgenite can be obtained from 14.0 grams of PbO and 14.0 grams of NaCl in the presence of excess water and carbon dioxide.
To determine the number of grams of phosgenite (Pb2Cl2CO3) that can be obtained from 14.0 g of PbO and 14.0 g of NaCl in the presence of excess water and carbon dioxide, we need to use stoichiometry.
1. Write the balanced chemical equation for the reaction:
PbO + 2 NaCl + H2O + CO2 -> Pb2Cl2CO3 + 2 NaOH
2. Calculate the molar mass of PbO:
Pb -> 207.2 g/mol
O -> 16.0 g/mol
Molar mass of PbO = 207.2 g/mol + 16.0 g/mol = 223.2 g/mol
3. Calculate the number of moles of PbO:
Moles of PbO = mass / molar mass
Moles of PbO = 14.0 g / 223.2 g/mol = 0.0627 mol
4. Use stoichiometry to find the number of moles of phosgenite:
From the balanced equation, the mole ratio between PbO and Pb2Cl2CO3 is 1:1.
Therefore, there will be 0.0627 mol of Pb2Cl2CO3 obtained.
5. Calculate the molar mass of Pb2Cl2CO3:
Pb -> 207.2 g/mol
Cl -> 35.5 g/mol (x2 since there are 2 Cl atoms)
C -> 12.0 g/mol
O -> 16.0 g/mol (x3 since there are 3 O atoms)
Molar mass of Pb2Cl2CO3 = 207.2 g/mol + 71.0 g/mol + 12.0 g/mol + 48.0 g/mol = 338.2 g/mol
6. Calculate the mass of phosgenite:
Mass of Pb2Cl2CO3 = moles of Pb2Cl2CO3 * molar mass
Mass of Pb2Cl2CO3 = 0.0627 mol * 338.2 g/mol = 21.2 g
Therefore, approximately 21.2 grams of phosgenite can be obtained from 14.0 grams of PbO and 14.0 grams of NaCl in the presence of excess water and carbon dioxide.