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2 Al + 3Cl2 --Ã 2AlCl3 so 2 moles of Aluminium atoms react with 3 moles of chlorine molecules or 1 atom of Al reacts with 3 atoms of chlorine. Aluminium has an atomic weight of 27 and Chlorine atoms 35.5. Multiplying by Avagadro's number 1 mole of Al atoms reacts with 3 moles of chlorine atoms. So 27 g of Al reacts with 106.5 g Chlorine (3 x 35.5) to produce 133.5 g AlCl3 15g Al = 15/27 = 0.5556 moles (4 decimal places) 2g Cl = 2/35.5 = 0.0563 moles of chlorine atoms but 3 needed per molecule of AlCl3 so 0.0563/3 = 0.0188 (4 decimal places) maximum number of moles of product possible. 0.0188 x 133.5 = 2.51 g maximum yield.
1 mole = 6.022e23 atoms 8.25 mole = 4.96815 e 24 atoms
C + 2Cl2 ==> CCl4atoms of C needed = 2.00 mol Cl2 x 1 mole C/2 mole Cl2 x 6.02x10^23 atoms/mole = 6.02x10^23 atoms
1 mol of anything = 6.022x1023 atomsso 1.4 mole = 1.4 * 6.022x1023 atoms / mole = 8.4308x1023 atoms
If you product has for example, dissociation of chlorine, it will have 2 atoms of chlorine in atomization, 0.5chlorine on balancing will give you only 1 mole of chlorine atom not 2 moles of atoms like dissociation enthalpy.
Molecular mass of AlCl3 = 27.0 + 3(35.5) = 133.5 Amount of AlCl3 = 400/133.5 = 3.00mol The ratio of Cl to AlCl3 is 3:1. Amount of Cl = 3 x 3.00 = 9.00mol
Avogadro's numbers worth. I mole of anything is, 6.022 X 1023 atoms ----------------------------
Six moles of HCl will be required: Each mole of chlorine contains two chlorine atoms, but each mole of HCl contains only one chlorine atom and the other reagent noted contains no chlorine atoms.
602200000000000000000000 atoms or 6.022 X 10^23
The answer depends on whether chlorine element or chlorine molecule is considered. 1 mole of Cl2 gas (chlorine molecule) has 6.023 x 1023 molecules or 12.046 x 1023 atoms. 1 mole of Cl element has 6.023 x 1023 atoms.
Chlorine gas is a diatomic molecule, Cl2. Two atoms per molecule. A mole contains Avogadro's Number of particles, that being (approximately) 6.02 x 1023. Hence, the number of atoms in one mole of chlorine gas is twice Avogadro's Number, or approximately 1.204 x 1024.
Chlorine gas is Cl2If you have 4.37 x 1018 atoms, you will have half that number of chlorine molecules.4.37 x 1018 atoms of Cl x 0.5 molecules of Cl2 / atoms of Cl = 2.20 x 1018 molecules of Cl2There are 6.02 x 1023 'things' in a mole.2.20 x 1018 molecules of Cl ÷ 6.02 x 1023 molecules / mole = 3.65 x 10-6 mole of Cl2.
10 grams of AlCl3 contains a certain number of moles in particles.1 mole of AlCl3 weighs (26.9815386+(3*35.453)=133.3405386), let's round that to 133.3405 grams.So 10 grams = 10/133.3405 = 0.075 mole of AlCl3 (rounded), but for each AlCl3 we have, you get 4 ions (1 Al3+, 3Cl-). So we have 0.075 moles, but we get 4 times as many ions. 0.075*4=0.3 mole1 mole = 6.022141x1023 (Avogadro's Number)0.3 mole = 1.81x1023 ions
It does NOT form molecules. When AlCl3 'breakdown ' it form the IONS Al^(3+) & 3 Cl^(-) AlCl3(s) IS a molecule.
lattice energy
2 Al + 3Cl2 --Ã 2AlCl3 so 2 moles of Aluminium atoms react with 3 moles of chlorine molecules or 1 atom of Al reacts with 3 atoms of chlorine. Aluminium has an atomic weight of 27 and Chlorine atoms 35.5. Multiplying by Avagadro's number 1 mole of Al atoms reacts with 3 moles of chlorine atoms. So 27 g of Al reacts with 106.5 g Chlorine (3 x 35.5) to produce 133.5 g AlCl3 15g Al = 15/27 = 0.5556 moles (4 decimal places) 2g Cl = 2/35.5 = 0.0563 moles of chlorine atoms but 3 needed per molecule of AlCl3 so 0.0563/3 = 0.0188 (4 decimal places) maximum number of moles of product possible. 0.0188 x 133.5 = 2.51 g maximum yield.
6.022 x 1023 molecules. This is Avogadro's number and a mole of anything. Since there are three atoms in each molecule, one of Barium and two of Chlorine, the number of atoms is three times that, or 1.807 x 1024.