if a student starts with 3.000 g of lead(II) carbonate for the first reaction and all other reagents are added in excess, what is the theoretical yield of lead(II) chloride solid?
lead is what is important here, so figure out how many moles of PbCO3 you have.
You will have that many moles of PbCl2
convert moles back to grams
To determine the theoretical yield of lead(II) chloride solid, we need to calculate the molar ratio between lead(II) carbonate and lead(II) chloride, and then use this ratio to calculate the amount of lead(II) chloride formed.
The balanced chemical equation for the reaction between lead(II) carbonate and hydrochloric acid (to form lead(II) chloride) is:
PbCO3 + 2HCl → PbCl2 + CO2 + H2O
From the equation, we can see that for every 1 mole of lead(II) carbonate (PbCO3) reacted, 1 mole of lead(II) chloride (PbCl2) is formed.
Step 1: Calculate the molar mass of lead(II) carbonate (PbCO3)
Pb: 207.2 g/mol
C: 12.0 g/mol
O: 16.0 g/mol (3 oxygen atoms)
Molar mass of PbCO3 = (207.2) + (12.0) + (16.0 * 3) = 267.2 g/mol
Step 2: Calculate the number of moles of lead(II) carbonate
Number of moles = mass / molar mass
Given: Mass = 3.000 g
Number of moles = 3.000 g / 267.2 g/mol = 0.0112 mol
Step 3: Calculate the number of moles of lead(II) chloride
Since the molar ratio between lead(II) carbonate and lead(II) chloride is 1:1, the number of moles of lead(II) chloride is also 0.0112 mol.
Step 4: Calculate the mass of lead(II) chloride
Mass = moles * molar mass
Molar mass of PbCl2 = (207.2) + (35.5 * 2) = 278.2 g/mol
Mass of PbCl2 = 0.0112 mol * 278.2 g/mol = 3.11 g
Therefore, the theoretical yield of lead(II) chloride solid is 3.11 grams.
To calculate the theoretical yield of lead(II) chloride (PbCl2) solid, we need to determine the balanced chemical equation for the reaction and the stoichiometry between the reactants and the product.
The balanced chemical equation for the reaction between lead(II) carbonate (PbCO3) and hydrochloric acid (HCl) to produce lead(II) chloride and carbon dioxide is:
PbCO3 + 2HCl -> PbCl2 + CO2 + H2O
From the balanced equation, we can see that 1 mole of lead(II) carbonate reacts with 2 moles of hydrochloric acid to produce 1 mole of lead(II) chloride.
To find the theoretical yield, we can use the molar mass of PbCO3 and the stoichiometry of the reaction.
1. Calculate the molar mass of PbCO3 (lead(II) carbonate):
The molar mass of PbCO3 is calculated by summing the atomic masses of each element:
Pb (lead) = 207.2 g/mol
C (carbon) = 12.01 g/mol
O (oxygen) = 16.00 g/mol (3 atoms)
Total molar mass = (207.2 g/mol) + (12.01 g/mol) + (16.00 g/mol x 3) = 327.21 g/mol
2. Convert the given mass of PbCO3 to moles:
Given mass = 3.000 g
Number of moles = (Given mass) / (Molar mass) = (3.000 g) / (327.21 g/mol) = 0.00917 mol
3. Determine the stoichiometry of the reaction:
Since the stoichiometry between PbCO3 and PbCl2 is 1:1, the number of moles of PbCO3 is equal to the number of moles of PbCl2.
4. Calculate the theoretical yield of PbCl2 in grams:
Theoretical yield = (Number of moles) x (Molar mass of PbCl2)
The molar mass of PbCl2 (lead(II) chloride) is calculated by summing the atomic masses of each element:
Pb (lead) = 207.2 g/mol
Cl (chlorine) = 35.45 g/mol (2 atoms)
Total molar mass = (207.2 g/mol) + (35.45 g/mol x 2) = 278.1 g/mol
Theoretical yield = (0.00917 mol) x (278.1 g/mol) = 2.55 g
Therefore, the theoretical yield of lead(II) chloride solid is 2.55 grams.