If the reaction were 2AlCl3 + 3Pb ---> 3PbCl2 + 2Al, then from this balanced equation, 14 moles of AlCl3 would produce 3/2 x 14 moles of PbCl2 = 21 moles of PbCl2.
21 moles of lead chloride are produced
lead (IV) oxide
-1. However PbCl is not a compound of lead- PbCl2 and PbCl4 are
Equilibrium is pushed toward the left in the equation. PbCl2(s) <=> Pb^2+(aq)+2Cl^-(aq).
lead (IV) oxide
-1. However PbCl is not a compound of lead- PbCl2 and PbCl4 are
Examples are: PbO, PbNO3, PbCl, PbS, etc.
Torstein Arnfinn Utigard has written: 'Wetting behaviour of molten PbCl r alkali chloride mixtures'
Equilibrium is pushed toward the left in the equation. PbCl2(s) <=> Pb^2+(aq)+2Cl^-(aq).
umber of Moles= Molar Mass (in g/mol) Mass (in grams) First, you'll need to know the molar mass of the substance in question. Here are the calculations for each sample: 6.684e13 pg of fluorine atoms: Molar mass of fluorine (F) = 19 g/mol Mass in grams = 6.684e13 pg = 6.684e-6 g Number of moles = (6.684e-6 g) / (19 g/mol) ≈ 3.52e-7 moles 2.435e6 mg of magnesium: Molar mass of magnesium (Mg) = 24.305 g/mol Mass in grams = 2.435e6 mg = 2.435 g Number of moles = (2.435 g) / (24.305 g/mol) ≈ 0.1001 moles 3.2e-3 kg of lead(II) chloride: Molar mass of lead(II) chloride (PbCl₂) = 207.2 g/mol Mass in grams = 3.2e-3 kg = 3200 g Number of moles = (3200 g) / (207.2 g/mol) ≈ 15.45 moles 6.684e-5 Mg of fluorine: Molar mass of fluorine (F) = 19 g/mol Mass in grams = 6.684e-5 Mg = 6.684e-5 g Number of moles = (6.684e-5 g) / (19 g/mol) ≈ 3.52e-6 moles 2.31e-9 Gg of carbon disulfide: Molar mass of carbon disulfide (CS₂) = 76.143 g/mol Mass in grams = 2.31e-9 Gg = 2.31e15 g Number of moles = (2.31e15 g) / (76.143 g/mol) ≈ 3.03e12 moles 4.91e9 ng of aluminum sulfate: Molar mass of aluminum sulfate (Al₂(SO₄)₃) = 342.15 g/mol Mass in grams = 4.91e9 ng = 4.91e-6 g Number of moles = (4.91e-6 g) / (342.15 g/mol) ≈ 1.43e-8 moles