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
To determine the number of grams of phosgenite that can be obtained, we need to calculate the limiting reactant in the given reaction. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
The balanced chemical equation for the reaction between PbO and NaCl in the presence of water and carbon dioxide is:
PbO + 2 NaCl + H2O + CO2 -> PbCO3 + 2 NaOH
First, we need to calculate the number of moles for each reactant. To do this, we divide the given mass by the molar mass of each compound.
Molar mass of PbO (lead(II) oxide) = 207.2 g/mol
Molar mass of NaCl (sodium chloride) = 58.44 g/mol
Number of moles of PbO = mass / molar mass = 10.0 g / 207.2 g/mol = 0.048 mol
Number of moles of NaCl = mass / molar mass = 10.0 g / 58.44 g/mol = 0.171 mol
Next, we compare the ratio of the coefficients in the balanced equation to determine the stoichiometry.
From the balanced equation, we can see that the stoichiometric ratio between PbO and PbCO3 is 1:1. This means that 1 mole of PbO reacts to produce 1 mole of PbCO3.
Since the stoichiometric ratio between NaCl and PbCO3 is 2:1, 2 moles of NaCl react to produce 1 mole of PbCO3.
Therefore, for the given amounts of PbO and NaCl, PbO is the limiting reactant because it produces the least amount of PbCO3.
To calculate the mass of phosgenite (PbCO3), we need to convert the number of moles of PbO to mass using the molar mass of PbCO3.
Molar mass of PbCO3 (phosgenite) = 267.2 g/mol
Mass of PbCO3 = moles of PbO * molar mass of PbCO3
= 0.048 mol * 267.2 g/mol
= 12.86 g
Therefore, the maximum amount of phosgenite that can be obtained from 10.0 g of PbO and 10.0 g of NaCl is 12.86 grams.