Answer:
I am not a man or boy, I am a girl so therefore I could answer this question before November ends Imao, have a great rest of your day.
Explanation:
Reduced- in a way losing an H means gaining an e-
C funnel because the funnel would have a large enough entrance to put a solid through
It will result in an increase in the rate of rxn
<span>atomic weights: Al = 26.98, Cl = 35.45
In this reaction; 2Al = 53.96 and 3Cl2 = 212.7
Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed.
Step 2:
(a) Ratio of Al:Cl = 2.70/4.05 = 0.6667
since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537.
so Cl is limiting
(b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced.
From Step 1:
212.7g of Cl will produce 266.66g AlCl3
212.7g = 266.66g
4.05g = x
x = 5.08g of AlCl3 can be produced
(c)
Al:Cl = 0.2537
Al:Cl = Al:4.05 = 0.2537
mass of Al used in reaction = 4.05 x 0.2537 = 1.027g
Excess reactant = 2.70 - 1.027 = 1.67g
King Leo · 9 years ago</span>
