Help plsss - Potassium iodide and lead(II) nitrate solutions react together to form solid lead(II) iodide and aqueous potassium nitrate: 2Kl(aq) + Pb(NO3)2(aq) → Pbl (s) + 2KNO3(aq)
In each of the following cases, carry out the calculations to determine the quantities required.
a) If 1.0 mol of potassium iodide reacts with 1.0 mol of lead(II) nitrate, determine which reactant is in
excess and by how many mol.
b) If 0.50 mol of potassium iodide reacts with 2.0 mol of lead(II) nitrate, determine which reactant is in
excess and by how many mol.
c) If 1.00g of lead(II) nitrate reacts with 1.50g of potassium iodide, determine which reactant is in
excess and the mass of lead(II) iodide that forms.

solute : a solute is a substance that is dissolved in another substance is known as a solvent. it is mixed with a special mixture know as homogeneous mixture.

solvent: a solvent is a substance that dissolved a solute, which results in a solution is known as solvent.

i believe the answer is three water molecules

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.