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poizon [28]
3 years ago
14

Determine the maximum number of moles of product that can be produced from 7.0 mol al and 8.0 mol cl2 according to the equation

2al + 3cl2 sa002-1.jpg 2 alcl3. describe in words the method used. then show the calculation(s).
Chemistry
1 answer:
Vilka [71]3 years ago
5 0
The balanced chemical reaction is

<span>2al + 3cl2 = 2alcl3

To determine the maximum amount of product, we need to determine which is the limiting reactant. Then, use the initial amount of that reactant to calculate the amount of the product that would be produced. We do as follows:

7 mol Al (3 mol Cl2 / 2 mol Al) = 10.5 mol Cl2
8 mol Cl2 ( 2 mol Al / 3 mol Cl2) = 5.3 mol Al
Thus, it is Cl2 that is the limiting reactant.

8 mol Cl2 ( 2 mol AlCl3 / 3 mol Cl2) = 5.3 moles of AlCl3 is produced</span>
You might be interested in
The chemical formula for ferric sulfate is Fe(SO4)3. Determine the following:
Lyrx [107]

Answer :

(a) The number of sulfur atoms are, 31.61\times 10^{23}.

(b) The mass of the mass of Fe_2(SO_4)_3 is, 1059.682 grams.

(c) The number of moles of Fe_2(SO_4)_3 is, 8.63\times 10^{-3}mole

(d) The mass of the mass of Fe_2(SO_4)_3 is, 19.95\times 10^{-22}g

Explanation :

(a) As we are given the number of moles of Fe_2(SO_4)_3 is, 1.75 mole. Now we have to calculate the number of sulfur atoms.

In the Fe_2(SO_4)_3, there are 2 iron atoms, 3 sulfur atoms, 12 oxygen atoms.

As, 1 mole of Fe_2(SO_4)_3 contains 3\times 6.022\times 10^{23} number of sulfur atoms.

So, 1.75 mole of Fe_2(SO_4)_3 contains 1.75\times 3\times 6.022\times 10^{23}=31.61\times 10^{23} number of sulfur atoms.

The number of sulfur atoms are, 31.61\times 10^{23}

(b) As we are given the number of moles of Fe_2(SO_4)_3 is, 2.65 mole. Now we have to calculate the mass of Fe_2(SO_4)_3.

\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3

The molar mass of Fe_2(SO_4)_3 = 399.88 g/mole

\text{Mass of }Fe_2(SO_4)_3=2.65mole\times 399.88g/mole=1059.682g

The mass of the mass of Fe_2(SO_4)_3 is, 1059.682 grams.

(c) As we are given the mass of Fe_2(SO_4)_3 is, 3.45 grams. Now we have to calculate the moles of Fe_2(SO_4)_3.

\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3

The molar mass of Fe_2(SO_4)_3 = 399.88 g/mole

3.45g=\text{Moles of }Fe_2(SO_4)_3\times 399.88g/mole

\text{Moles of }Fe_2(SO_4)_3=8.63\times 10^{-3}mole

The number of moles of Fe_2(SO_4)_3 is, 8.63\times 10^{-3}mole

(d) As we are given the formula unit of Fe_2(SO_4)_3 is, 3. Now we have to calculate the mass of Fe_2(SO_4)_3.

As we know that 1 mole of Fe_2(SO_4)_3 contains 6.022\times 10^{23} formula unit.

Formula used :

\text{Formula unit of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times 6.022\times 10^{23}

3=\text{Moles of }Fe_2(SO_4)_3\times 6.022\times 10^{23}

\text{Moles of }Fe_2(SO_4)_3=4.989\times 10^{-24}mole

Now we have to calculate the mass of Fe_2(SO_4)_3.

\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3

The molar mass of Fe_2(SO_4)_3 = 399.88 g/mole

\text{Mass of }Fe_2(SO_4)_3=4.989\times 10^{-24}mole\times 399.88g/mole=19.95\times 10^{-22}g

The mass of the mass of Fe_2(SO_4)_3 is, 19.95\times 10^{-22}g

6 0
3 years ago
If you test the boiling point of ethanol and have a 2.3% error, what was the boiling point of ethanol in the test?
anyanavicka [17]

Answer:

the answer is A

Explanation:

3 0
2 years ago
2. if 0.20 m fe3 had been used instead of 0.020 m fe3 , how would the numerical value of the rate constant and the activation en
dezoksy [38]

the calculated value is Ea is 18.2 KJ and A is 12.27.

According to the exponential part in the Arrhenius equation, a reaction's rate constant rises exponentially as the activation energy falls. The rate also grows exponentially because the rate of a reaction is precisely proportional to its rate constant.

At 500K, K=0.02s−1

At 700K, k=0.07s −1

The Arrhenius equation can be used to calculate Ea and A.

RT=k=Ae Ea

lnk=lnA+(RT−Ea)

At 500 K,

ln0.02=lnA+500R−Ea

500R Ea (1) At 700K lnA=ln (0.02) + 500R

lnA = ln (0.07) + 700REa (2)

Adding (1) to (2)

700REa100R1[5Ea-7Ea] = 0.02) +500REa=0.07) +700REa.

=ln [0.02/0 .07]

Ea= 2/35×100×8.314×1.2528

Ea =18227.6J

Ea =18.2KJ

Changing the value of E an in (1),

lnA=0.02) + 500×8.314/18227.6

= (−3.9120) +4.3848

lnA=0.4728

logA=1.0889

A=antilog (1.0889)

A=12.27

Consequently, Ea is 18.2 KJ and A is 12.27.

Learn more about Arrhenius equation here-

brainly.com/question/12907018

#SPJ4

5 0
1 year ago
How many grams of sodium are needed to produce 12.5g of sodium oxide
Hatshy [7]

Answer:

25 possibly

Explanation:

I'm not too sure about this, but sodium oxide is Na2O, 2 sodium and 1 oxygen, so 12.5g * 2 is 25

If someone else comes up with a more convincing argument listen to them

4 0
3 years ago
a chemist determined by measurements that 0.015 moles of tungsten participated in a chemical reaction. Calculate the mass of tun
ICE Princess25 [194]

Answer: 0.84g

Explanation:

8 0
3 years ago
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