How many grams of phosgenite can be obtained from 10.0 g of PbO and 10.0 g of NaCl in the presence of excess water and carbon dioxide
Do you have the equation?
unbalanced:
PbO+NaCl+H2O+CO2 yields Pb2Cl2CO3+NaOH
Balance the equation.
2PbO + 2NaCl + H2O + CO2 ==> (PbCl)2CO3 + 2NaOH. Check that to mae sure.
mols PbO = gram/molar mass
mols NaCl = grams/molar mass.
From PbO, convert to mols product.
That's mols PbO from above x (1/2) = ?
From NaCl, convert to mols product.
That's mols NaCl x 1/2 = ?
Most likely these two numbers will not be the same; the correct value in limiting reagent problems (which this is) is always the smaller value (and the reagent producing that value is the limiting reagent).
Take the smaller value and convert to grams product.
That's mols product x molar mass product = grams product.
To determine the number of grams of phosgenite (Pb2Cl2CO3) that can be obtained from 10.0 g of PbO and 10.0 g of NaCl, we need to calculate the limiting reactant and then use stoichiometry to find the corresponding amount of phosgenite produced.
Here's how you can solve it step by step:
Step 1: Write the balanced chemical equation for the reaction:
2PbO + 2NaCl + H2O + CO2 -> Pb2Cl2CO3 + 2NaOH
Step 2: Calculate the molar masses of PbO, NaCl, and Pb2Cl2CO3:
Molar mass of PbO = 207.2 g/mol
Molar mass of NaCl = 58.44 g/mol
Molar mass of Pb2Cl2CO3 = 448.6 g/mol
Step 3: Convert the given masses of PbO and NaCl to moles:
Moles of PbO = mass / molar mass = 10.0 g / 207.2 g/mol = 0.0482 mol
Moles of NaCl = mass / molar mass = 10.0 g / 58.44 g/mol = 0.171 mol
Step 4: Determine the limiting reactant. The reactant that produces the least amount of product is the limiting reactant. To do this, we need to compare the moles of each reactant to the stoichiometry of the balanced equation:
From the balanced equation, the stoichiometry is 2 moles of PbO reacts with 2 moles of NaCl to produce 1 mole of Pb2Cl2CO3.
Since the mole ratio of PbO to NaCl is 2:2, the amount of NaCl required to react with the given amount of PbO is also 0.0482 mol (0.171 mol × (2 moles NaCl / 2 moles PbO)).
Since we have an equal amount of moles for both reactants, PbO and NaCl, the limiting reactant is PbO.
Step 5: Calculate the amount of phosgenite (Pb2Cl2CO3) that can be obtained from the limiting reactant:
The mole ratio between PbO and Pb2Cl2CO3 is 2:1 (from the balanced equation). Therefore, 0.0482 mol of PbO will react to produce 0.0241 mol of Pb2Cl2CO3 (0.0482 mol × (1 mole Pb2Cl2CO3 / 2 moles PbO)).
Step 6: Convert the moles of Pb2Cl2CO3 to grams:
Grams of Pb2Cl2CO3 = moles × molar mass = 0.0241 mol × 448.6 g/mol = 10.82 g
Therefore, you can obtain approximately 10.82 grams of phosgenite (Pb2Cl2CO3) from 10.0 grams of PbO and 10.0 grams of NaCl in the presence of excess water and carbon dioxide.