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3 years ago

If 3.3167 moles of Al are produced, how many moles of AlCl3 were reacted? I

Chemistry

1 answer:

tresset_1 [31]3 years ago

8 0

Answer:

3.3167 moles Of AlCl3

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3Ca + 2AlCl3 —> 3CaCl2 + 2Al

From the balanced equation above,

2 moles of AlCl3 reacted to produce 2 moles of Al.

Finally, we shall obtained the number of moles of AlCl3 that reacted to produce 3.3167 moles of Al as follow:

From the balanced equation above,

2 moles of AlCl3 reacted to produce 2 moles of Al.

Therefore, 3.3167 moles Of AlCl3 will also react to produce 3.3167 moles of Al.

Thus, 3.3167 moles Of AlCl3 is needed for the reaction.

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Elements in the same period share the same number of:

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2 years ago

2. Consider the reaction 2 Cg H18 (4) +250â (9) âº 16 co, (g) + 18 HâO(g) la How many moles of H20co) are produced, when |--16:1

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Answer :

(a) The moles of water produced are 145.35 moles.

(b) The mass of oxygen needed are 3080.8 grams.

Solution for part (a) : Given,

Moles of $C_8H_{18}$ = 16.15 moles

First we have to calculate the moles of $H_2O$

The balanced chemical reaction is,

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_8H_{18}$ react to give 18 moles of $H_2O$

So, 16.15 moles of $C_8H_{18}$ react to give $\frac{16.15}{2}\times 18=145.35$ moles of $H_2O$

The moles of water produced are 145.35 moles.

Solution for part (b) : Given,

Mass of $C_8H_{18}$ = 878 g

Molar mass of $C_8H_{18}$ = 114 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_8H_{18}$ .

$\text{ Moles of }C_8H_{18}=\frac{\text{ Mass of }C_8H_{18}}{\text{ Molar mass of }C_8H_{18}}=\frac{878g}{114g/mole}=7.702moles$

Now we have to calculate the moles of $O_2$

The balanced chemical reaction is,

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_8H_{18}$ react with 25 moles of $O_2$

So, 7.702 moles of $C_8H_{18}$ react with $\frac{7.702}{2}\times 25=96.275$ moles of $O_2$

Now we have to calculate the mass of $O_2$ .

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(96.275moles)\times (32g/mole)=3080.8g$

The mass of oxygen needed are 3080.8 grams.

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3 years ago

1. Show the mechanism for the main product for the monochlorination of 2-methylbutane. List all other products.

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Answer:

See explanation below

Explanation:

First, we need to understand that the monochlorination of an alkane like this one, involves substitution of one of the atoms of hydrogen of the molecule for an atom of chlorine.

This reaction takes place when the alkane reacts with Cl₂ in presence of light or heat.

When this happens, the first step involves the breaking of the double bond of the chlorine to form the ion Cl⁻.

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In the picture attached you have the mechanism and product for this reaction:

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3 years ago

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