How many grams of the excess reactant are left over when 4.43E-1 moles of Al(s) react with 1.29E+0 moles of Cl(g)? 2Al(s) + 3Cl(g) → 2AlCl(s)

44.41g

Explanation:

2Al + 3Cl2 —> 2AlCl3

From the equation, 2 moles of Al required 3 moles.

From the question,

Number of mole of Al = 4.43E-1 = 4.43x10^-1 = 0.443mol

Number of mole of Cl = 1.29E+0 = 1.29x10^0 = 1.29mol

Let's try something out:

First let us used the number of mole of Al given from the question to detect which is excess

2moles of Al required 3 moles of Cl

0.443mol of Al will require = (0.443x3) / 2 = 0.6645mol of Cl

Comparing this mol of Cl2 obtained, with what is given(1.29mol) we see that Cl is excess.

Now let us use the number of mole of Cl2 given from the question:

2 moles of Al required 3 moles of Cl2.

Therefore Xmol of Al will require 1.29mol of Cl i.e

Xmol of Al = (2x1.29) / 3 = 0.86mol of Al.

Comparing this mol of Al obtained with what is given (0.443mol) we see that Al is the limiting reactant.

Now we know that Cl is excess

The excess Cl = 1.29 — 0.6645 = 0.6255mol

Converting the excess 0.6255mol of Cl to gram, we have;

Mass = n x molar Mass

n = 0.6255

Molar Mass of Cl2 = 2 x35. 5 = 71g/mol

Mass = 0.6255 x 71 = 44.41g

i think its 100

i hope this !