3Pb(NO3)2 + 2AlCl3 = 3PbCl2 + 2Al(NO3)3 Is the balanced equation.
If there are 8 moles of ALCl3 used instead of 2, the number of moles need of the first reactant needed will be 6 moles.
The new balanced equation is: 6Pb(NO3)2 + 8AlCl3 = 6PbCl2 + 8Al(NO3)3
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.
It does NOT form molecules. When AlCl3 'breakdown ' it form the IONS Al^(3+) & 3 Cl^(-) AlCl3(s) IS a molecule.
Balanced equation: 2Al + 3Cl2 -> 2AlCl3For every 3 moles of Cl2, 2 moles of AlCl3 is produced (Using the numbers in front of the compounds)Now set up a proportion: 3/2 = 0.30/?Cross Multiply: (2 X 0.30) / 3 = 0.20.2 moles of AlCl3 will be produced.
The answer is 144,007 g for the anhydrous aluminium chloride.
0.355 M AlCl3 (3 moles Cl/1 mole AlCl3)= 1.07 M Cl================Naturally.0.355 M Al============add1.43 M total=========
0,75 moles of AlCl3 (anhydrous) is equivalent to 100,005 g.
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/liter * .0655 L = 0.013755 moles AlCl3 .013755 * 3 moles Cl / 1 mole AlCl3 = 0.041265 moles Cl * avagadro's number = number of chloride ions
3Pb(NO3)2 + 2AlCl3 = 3PbCl2 + 2Al(NO3)3 Is the balanced equation.If there are 8 moles of ALCl3 used instead of 2, the number of moles need of the first reactant needed will be 6 moles.The new balanced equation is: 6Pb(NO3)2 + 8AlCl3 = 6PbCl2 + 8Al(NO3)3
It does NOT form molecules. When AlCl3 'breakdown ' it form the IONS Al^(3+) & 3 Cl^(-) AlCl3(s) IS a molecule.
Equation. 2Al + 3Cl2 -> 2AlCl3 one to one again 0.440 moles Al (2 moles AlCl3/2 moles Al) = 0.440 moles AlCl3 produced
Balanced equation: 2Al + 3Cl2 -> 2AlCl3For every 3 moles of Cl2, 2 moles of AlCl3 is produced (Using the numbers in front of the compounds)Now set up a proportion: 3/2 = 0.30/?Cross Multiply: (2 X 0.30) / 3 = 0.20.2 moles of AlCl3 will be produced.
Al+HCl===> AlCl3+H2 Is the reaction. You need &.2 moles of HCl.
6.75 mol AlCl3 x 6.022 x 10^23 formula units ------------------------------------- = Your Answer 1 mol AlCl3
The mass of aluminium is 11,2 g.
0.2550 g AlC3 (1 mol/132 g) =0.001932 mol AlCl3 0.001932 mol AlCl3 (6.022 x 10^23 molecules AlCl3/1 mol AlCl3) = 1.163 x 10^21 1.163x10^21 molecules AlCl3 (3 mol Cl/1 mol AlCl3) =3.490x10^21 Cl ions 3.490x10^21 Cl ions (1 mol/6.022 x 10^23) =5.795x10^-3 moles Cl The formula to solve this problem appears above.
The answer is 144,007 g for the anhydrous aluminium chloride.