If 3.3167 moles of Al are produced, how many moles of
AlCl3 were reacted?
I

3.3167 moles Of AlCl3

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3Ca + 2AlCl3 —> 3CaCl2 + 2Al

From the balanced equation above,

2 moles of AlCl3 reacted to produce 2 moles of Al.

Finally, we shall obtained the number of moles of AlCl3 that reacted to produce 3.3167 moles of Al as follow:

From the balanced equation above,

2 moles of AlCl3 reacted to produce 2 moles of Al.

Therefore, 3.3167 moles Of AlCl3 will also react to produce 3.3167 moles of Al.

Thus, 3.3167 moles Of AlCl3 is needed for the reaction.

mechanics (greek   is that area of science concerned with the behaviour of physical bodies when subjected to forces or displacements, and the subsequent effects of the bodies on their environment. the scientific discipline has its origins in ancient greece with the writings of aristotle and archimedes[1][2][3] (see history of classical mechanics and timeline of classical mechanics). during the early modern period, scientists such as galileo, kepler, and newton laid the foundation for what is now known as classical mechanics. it is a branch of classical physics that deals with particles that are either at rest or are moving with velocities significantly less than the speed of light. it can also be defined as a branch of science which deals with the motion of and forces on objects. the field is yet less widely understood in terms of quantum theory.

explanation:

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g