<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>
Answer:
A = 2A + 3B → 5C
Explanation:
The two molecule of A and three molecules of B will react to form the five molecules of C.
2A + 3B → 5C
Other options are incorrect because,
B = A₂ + B₃ → C₅
in this reaction one molecule of A₂ and one molecule of B₃ combine to form one molecule of C₅.
C = 2A + 5B → 3C
in this reaction two molecules of A and five molecules of B combine to form three molecule of C.
D = A₂ + B₃ → C₃
in this reaction one molecule of A₂ and one molecule of B₃ combine to from one molecule of C₃.
Answer:
the equator is closer to the sun
