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vova2212 [387]
3 years ago
7

Consider the balanced equation below.

Chemistry
1 answer:
pochemuha3 years ago
8 0

A. 1:3

The equation shows that 1 mol Fe_3_O4 ≡ 3 mol Fe.

∴ The mole ratio is Fe_3_O4:Fe = 1:3.

“B. 1:4” is <em>incorrect</em>. It is the mole ratio of <em>Fe_3O_4:H_2O </em>and of<em> Fe_3O_4:H_2</em>.

“C: 3:1” is <em>incorrect</em>. It is the mole ratio of <em>Fe:Fe_3O_4</em>.

“D: 4:3” is <em>incorrect</em>. It is the mole ratio of <em>H_2O:Fe and of H_2:Fe</em>.

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2. Consider the reaction 2 Cg H18 (4) +250â (9) ⺠16 co, (g) + 18 HâO(g) la How many moles of H20co) are produced, when |--16:1
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(a) The moles of water produced are 145.35 moles.

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

<u>Solution for part (a) : Given,</u>

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.

<u>Solution for part (b) : Given,</u>

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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