Be sure to answer all parts. The balanced equation for the reaction of aluminum metal and chlorine gas is 2Al(s) + 3Cl2(g) → 2AlCl3(s) Assume that 0.80 g Al is mixed with 0.23 g Cl2. (a) What is the limiting reactant? Cl2 Al (b) What is the maximum amount of AlCl3, in grams, that can be produced? g AlCl3

a) is the limiting reagent

b)  0.27 g of will be produced.

Explanation:

To calculate the moles :

According to stoichiometry :

3 moles of require = 2 moles of

Thus 0.003 moles of will require=  of

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

b) As 3 moles of give = 2 moles of

Thus 0.003 moles of give =  of

Mass of

Thus 0.27 g of will be produced.

it should be p1 v1 = p2 v2

4d^55s^1 ( third choice)