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How many moles of pb(no3)2 are required of 8 moles of alcl3 are consumed?
Ashlyn Coffield
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
Jarrod Ruecker
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How many moles of PbCl₂ are produced if 14 moles of AlCl₃ are consumed?
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.
How many molecules does alcl3 dissociate into?
It does NOT form molecules. When AlCl3 'breakdown ' it form the IONS Al^(3+) & 3 Cl^(-) AlCl3(s) IS a molecule.
How many moles of AlCl3 are produced from 0.30 mole of Cl2 Show the work?
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.
What is the mass in grams of 1.05 moles of AlCl3?
The answer is 144,007 g for the anhydrous aluminium chloride.
What is the total molar concentration of ions in an 0.355 M solution of AlCl3?
0.355 M AlCl3 (3 moles Cl/1 mole AlCl3)= 1.07 M Cl================Naturally.0.355 M Al============add1.43 M total=========
How many formula units are in 0.75 moles of AlCl3?
0,75 moles of AlCl3 (anhydrous) is equivalent to 100,005 g.
How many moles of PbCl₂ are produced if 14 moles of AlCl₃ are consumed?
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.
How many chloride ions are present in 65.5 mL of 0.210 M AlCl3 solution?
.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
How many moles of Pb(NO₃)₂ are required if 8 moles of AlCl₃ are consumed?
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
How many molecules does alcl3 dissociate into?
It does NOT form molecules. When AlCl3 'breakdown ' it form the IONS Al^(3+) & 3 Cl^(-) AlCl3(s) IS a molecule.
When 0.440 mol of aluminum are allowed to react with an excess of chlorine gas Cl2 how many moles of aluminum chloride are produced?
Equation. 2Al + 3Cl2 -> 2AlCl3 one to one again 0.440 moles Al (2 moles AlCl3/2 moles Al) = 0.440 moles AlCl3 produced
How many moles of AlCl3 are produced from 0.30 mole of Cl2 Show the work?
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.
How many moles of HCl and needed to react 2.4 moles of Al?
Al+HCl===> AlCl3+H2 Is the reaction. You need &.2 moles of HCl.
Calculate the number of formula units of the compound and the number of each ion in 6.75 moles of AlCl3?
6.75 mol AlCl3 x 6.022 x 10^23 formula units ------------------------------------- = Your Answer 1 mol AlCl3
How many grams Al from 0.4 moles AlCl3?
The mass of aluminium is 11,2 g.
How many moles of HCI do you need to react with aluminum foil if you use 15 grams of acid?
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.
What is the mass in grams of 1.05 moles of AlCl3?
The answer is 144,007 g for the anhydrous aluminium chloride.
