If you react 25.0g of Cu with 25.0g of AlCl3 in the following reaction 3Cy + 2AlCl3 -> 3CuCl2 + 2Al a. find the excess and limiting reactants
b. calculate the mass of leftover reactant

a. AlCl₃ ⇒ limiting reactant(smaller ratio)

Cu ⇒ excess reactant

b. the mass of leftover reactant : 7.207 g

Given

25 g Cu

25 g AlCl3

Required

a. the excess and limiting reactants

b.  the mass of leftover reactant

Solution

Reaction

3Cu + 2AlCl₃ ⇒ 3CuCl₂ + 2Al

mol Cu(Ar = 63.5 g/mol) :

mol = mass : Mw

mol = 25 : 63.5

mol = 0.394

mol AlCl3(MW=133,34 g/mol) :

mol = 25 : 133,34 g/mol

mol = 0.187

mol ratio to reaction coefficient Cu : AlCl₃ =

AlCl₃ ⇒ limiting reactant(smaller ratio)

Cu ⇒ excess reactant

b. the mass of leftover reactant :

mol Cu = 3/2 x 0.187 = 0.2805

mol left = 0.394 - 0.2805 = 0.1135

mass = 0.1135 x 63.5 = 7.207 g

the molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

explanation:

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.