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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
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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 limiting reagent of CuI2 AlCl3 3CuCl2 AlI3?
The balanced reaction of cupric iodide and aluminum chloride is 3 CuI2 + 2 AlCl3 = 3 CuCl2 + 2 AlI3. This means that the limiting reaction is cupric iodide, because more of it is required than aluminum chloride.
What is the mass in grams of 1.05 moles of AlCl3?
The answer is 144,007 g for the anhydrous aluminium chloride.
How many moles of PbCl₂ are produced if 14 moles of AlCl₃ are consumed?
Using the balanced equation 2 AlCl₃ + 3 Pb(NO₃)₂ → 3 PbCl₂ + 2 Al(NO₃)₃, the mole ratio between AlCl₃ and PbCl₂ is 2:3. Therefore, if 14 moles of AlCl₃ are consumed, 9.33 moles (14 moles / 2 * 3) of PbCl₂ will be produced.
What is the mass of AlCl3 that is produced when 25.0 grams of Al2O3 reacts with HCl according Al2O3 6HCl to 2ALCl3 3H2O?
To find the mass of AlCl3 produced, first calculate the molar mass of Al2O3 and AlCl3. Then, use stoichiometry to determine the moles of AlCl3 produced. Finally, convert moles of AlCl3 to grams using its molar mass. The mass of AlCl3 produced will be 12.25 grams.
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 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 chloride ions are present in 65.5 mL of 0.210 M AlCl3 solution?
In 1 mol of AlCl3, there are 3 chloride ions. First calculate the moles of AlCl3 in the solution: 65.5 mL is 0.0655 L. Multiply 0.0655 L by 0.210 mol/L to get the moles of AlCl3. Finally, multiply this by 3 to find the number of chloride ions in the solution.
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
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 limiting reagent of CuI2 AlCl3 3CuCl2 AlI3?
The balanced reaction of cupric iodide and aluminum chloride is 3 CuI2 + 2 AlCl3 = 3 CuCl2 + 2 AlI3. This means that the limiting reaction is cupric iodide, because more of it is required than aluminum chloride.
How many moles of PbCl₂ are produced if 14 moles of AlCl₃ are consumed?
Using the balanced equation 2 AlCl₃ + 3 Pb(NO₃)₂ → 3 PbCl₂ + 2 Al(NO₃)₃, the mole ratio between AlCl₃ and PbCl₂ is 2:3. Therefore, if 14 moles of AlCl₃ are consumed, 9.33 moles (14 moles / 2 * 3) of PbCl₂ will be produced.
What is the mass of AlCl3 that is produced when 25.0 grams of Al2O3 reacts with HCl according Al2O3 6HCl to 2ALCl3 3H2O?
To find the mass of AlCl3 produced, first calculate the molar mass of Al2O3 and AlCl3. Then, use stoichiometry to determine the moles of AlCl3 produced. Finally, convert moles of AlCl3 to grams using its molar mass. The mass of AlCl3 produced will be 12.25 grams.
How many grams of aluminum chloride are produced when 18 grams of aluminum are reacted with an excess of hydrochloride acid?
The empirical formula for aluminum chloride is AlCl3, and its gram formula mass is 133.34. The formula shows that each formula unit contains one aluminum atom, and the the gram atomic mass of aluminum is 26.9815. Therefore, 18(133.34/26.9815) or 89 grams, to the justified number of significant digits, of aluminum chloride 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.
How many grams Al from 0.4 moles AlCl3?
The mass of aluminium is 11,2 g.
Calculate the number of formula units of the compound and the number of each ion in 6.75 moles of AlCl3?
In 6.75 moles of AlCl3, there are 6.75 times Avogadro's number of formula units, which is approximately 4.07 x 10^24 formula units. Each formula unit of AlCl3 contains one Al^3+ ion and three Cl^- ions. Therefore, in 6.75 moles of AlCl3, there are also 4.07 x 10^24 Al^3+ ions and 1.22 x 10^25 Cl^- ions.
