Two students react 15.50g of Lead (II) nitrate with 3.81g of sodium chloride.
Predict the mass of the excess reactant left over when the reaction is complete.
Here is a worked example of a limiting reagent problem. At the end of the example it shows how to solve for the excess reagent. Post back if you have any questions about the example.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
Thank you very much
To predict the mass of the excess reactant left over when the reaction is complete, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in a chemical reaction, limiting the amount of product that can be formed.
To find the limiting reactant, we have to compare the moles of the two reactants. We can follow these steps:
1. Calculate the number of moles for each reactant:
- Moles of Lead (II) nitrate (Pb(NO3)2) = mass / molar mass
- Moles of sodium chloride (NaCl) = mass / molar mass
The molar mass of Lead (II) nitrate (Pb(NO3)2) is 331.2 g/mol, and the molar mass of sodium chloride (NaCl) is 58.44 g/mol.
2. Calculate the mole ratio between the reactants by dividing the number of moles of one reactant by the coefficient in the balanced chemical equation.
- The balanced chemical equation for the reaction between lead (II) nitrate and sodium chloride is:
Pb(NO3)2 + 2 NaCl -> PbCl2 + 2 NaNO3
3. Compare the mole ratio of the reactants. The reactant that has a smaller mole ratio is the limiting reactant.
- If the mole ratio of Pb(NO3)2 to NaCl is 1:2, then we need twice as many moles of NaCl to react with the available moles of Pb(NO3)2.
- If the mole ratio of Pb(NO3)2 to NaCl is greater than 1:2, then Pb(NO3)2 is the limiting reactant.
- If the mole ratio of Pb(NO3)2 to NaCl is less than 1:2, then NaCl is the limiting reactant.
4. Once we have determined the limiting reactant, we can calculate the amount of excess reactant remaining by subtracting the moles of the limiting reactant used from the total moles of the excess reactant.
- Moles of excess reactant = Moles of excess reactant - (Moles of limiting reactant * Excess reactant coefficient in the balanced equation)
Finally, use the moles of the excess reactant and its molar mass to calculate the mass of the excess reactant left over.