For the following chemical reaction, how many moles of lead(ii)iodide (PbI2) will be produced from 8 mol of potassium iodide (KI)?
To determine the moles of lead(II) iodide (PbI2) produced from 8 mol of potassium iodide (KI), we first need to know the balanced chemical equation for the reaction. Without the equation, we cannot determine the stoichiometry between the reactants and products.
Please provide the balanced chemical equation for the reaction.
To determine the number of moles of lead(II) iodide (PbI2) produced, we need to balance the chemical equation for the reaction between lead(II) nitrate (Pb(NO3)2) and potassium iodide (KI):
Pb(NO3)2 + 2KI → PbI2 + 2KNO3
From the balanced equation, we can see that 1 mole of Pb(NO3)2 reacts with 2 moles of KI to produce 1 mole of PbI2.
Therefore, we need to calculate the number of moles of PbI2 produced from 8 moles of KI using the stoichiometric ratio:
8 moles KI × (1 mole PbI2 / 2 moles KI) = 4 moles PbI2
So, 8 moles of potassium iodide (KI) will produce 4 moles of lead(II) iodide (PbI2) in this chemical reaction.
I don't see a chemical reaction so I'll assume one.
2KI + Pb^2+ ==> PbI2 + 2K^+.
8 mols KI x (1 mol PbI2/2 mols KI) = 8 x 1/2 = ?