For Part A: Find Limiting Reactant of 2.70g Al & 4.05g Cl2

[mugzieee]

2Al+3Cl2=Al2Cl6
a)which reactant is limiting if 2.70g Al and 4.05g Cl3 is mixed.
b)what mass of Al2Cl3 can be produced.

for part a i found the limiting reactant to be Cl2. I don't understand what I am supposed to do for the second part. Should i write another equation as Al2Cl3 as the product, Al, and Cl2 as the reactants, balance it, and then use the limiting reagant from the aboove part to find the mass of Al2Cl3? I am not really sure what the question is trying to ask.

 

[Nenad]

use mol ratio with your lim. reagant to find out how many mol of al2cl6 you will have. then multiply this out with the molar mass of the product.

regards,

Nenad

 

[Nenad]

ok, here is what I am getting for an answer. The total amount of product produced is 0.019mol and the mass of this is:
$$0.057mol Cl_{2} x \frac{1mol Al_{2}Cl_{6}}{3mol Cl_{2}} = 0.019mol Al_{2}Cl_{6}$$
$$0.019mol Al_{2}Cl_{6} x \frac{266.78g Al_{2}Cl_{6}}{1mol Al_{2}Cl_{6}} = 5.07g Al_{2}Cl_{6}$$

thats my answer, what are you getting?

Regards,

Nenad