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3 years ago

Balance the reaction and determine which of the following coefficients

Chemistry

1 answer:

tatiyna3 years ago

6 0

Answer:

B. The correct ratio is 3:1:1:3

Explanation:

According to this question, the following chemical equation was given:

CH3COOH + Al(OH)3 → Al(CH3COO)3 + H2O

This equation is obviously not balanced because the number of atoms of each element on both sides of the equation is not equal or the same. To balance the equation, we make use of coefficients as follows:

3CH3COOH + Al(OH)3 → Al(CH3COO)3 + 3H2O

- There are now 6 moles of C on each side

- There are now 15 moles of H on each side

- There are now 9 moles of O on each side

- There is now 1 mole of Al on each side

The correct ratio of each compound is in the order 3:1:1:3

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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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