Aluminum metal and chlorine gas react to from aluminum chloride 2al(s) + 3cl2(g) --> al2cl6(s) how many grams of al2cl6 can be produced from the reaction of 2.67 mol aluminum and 3.26 mol cl2?

290.1 grams.

Explanation:

2 Al + 3Cl2 ---> Al2Cl6

2 moles of Al react with 3 moles of Cl2 to form 1 mole of Al2Cl6

So 2.67 moles Al  reacts with 3/2 * 2.67 moles Cl2 = 4 moles of chlorine. But we only have 3.26 moles of Cl2 so this will react with only part of the available Al.

3.26 reacts with 2/3 * 3.26 = 2.173 moles  of aluminium.

So the  number of moles of Al2Cl6 that can be produced = 2.173 * 1/2

= 1.087 moles.

= 1.087 * (2*26.98 + 6*35.45)

= 290.1 grams.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

water is used to store them because it would keep the radiation from getting out in the open and causing harm to the workers