What is the limiting reactant when 415 mL of 3.0M copper(II) chloride solution is added to 25.0 g of solid aluminum?

The answer to your question is Copper (II) chloride

Explanation:

Data

Limiting reactant = ?

volume of CuCl₂ = 415 ml

[CuCl₂] = 3.0 M

mass of Al = 25 g

Process

1.- Calculate the mass of CuCl₂ in solution

Molarity = moles / volume

moles = Molarity x volume

moles = 3 x 0.415

moles = 1.245

Molar mass of CuCl₂ = 63.5 + (35.5 x 2)

                                  = 63.5 + 71

                                  = 134.5 g

                   134.5 g 1 mol

                      x         1.245 moles

                      x = (1.245 x 134.5)/1

                      x = 167.5 g of CuCl₂

2.- Balanced chemical reaction

                  3CuCl₂  +  2 Al   ⇒    2AlCl₃  +  3Cu

3.- Calculate proportions

Theoretical proportion = (3 x 134.5)/(2 x 27) = 403.5/54 = 7.47

Experimental proportion = 167.5 g / 25 g = 6.7

4.- Conclusion

As the experimental proportion was lower than the theoretical proportion the limiting reactant is CuCl₂.

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .