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dedylja [7]
3 years ago
9

The two balanced equations (1) and (2) are for reactions in which gaseous carbon dioxide is produced from the combustion of (1)

solid carbon and (2) gaseous carbon monoxide.
1. C(s) + O2(g) CO2(g) + 94.0 kcal
2. CO(g) + ½ O2(g) CO2(g) + 67.6 kcal
When 112 grams of carbon monoxide are consumed according to equation 2, which of the following occurs? (atomic weights: C = 12.0 g/mol, O = 16.0 g/mol)
1.0 mole of carbon dioxide is produced.
67.6 kcal of heat are generated.
2.0 moles of oxygen are consumed.
0.25 mole of carbon dioxide is produced.
0.50 mole of oxygen is consumed.
Chemistry
1 answer:
V125BC [204]3 years ago
3 0

Answer:

2.0 mol of oxygen are consumed.

Step-by-step explanation:

You know that you will need a balanced equation with masses, moles, and molar masses, so gather all the information in one place.

M_r:   28.0                 44.0

          CO  + ½O₂ ⟶ CO₂ + 67.6 kcal

m/g:   112

<em>Step 1</em>. Convert grams of CO to moles of CO

1 mol CO = 28.0 g CO

Moles of CO = 112 × 1/28.0

<em>Step 2.</em> Convert moles of CO to moles of CO₂.

The molar ratio is 1 mol CO₂ to 1 mol CO

Moles of CO₂ = 4.000 × 1/1

Moles of CO₂ = 4.00 mol CO₂

Option A is <em>wrong</em>.

<em>Step 3.</em> Calculate the amount of heat generated.

q = ΔH

The conversion factor is 67.6 kcal/1 mol CO₂

q = 4.00 × 67.6

q = 270 kJ

Option B is <em>wrong</em>, because it gives the heat generated by 1 mol of CO.

<em>Step 4. </em>Calculate the moles of O₂ consumed

Moles of O₂ = 2.00 mol O₂

Option C is correct.

<em>Step 5.</em> Calculate the moles of CO₂ formed

Already done in Step 2.

Moles of CO₂ = 4.00 mol CO₂

Option D is <em>wrong. </em>

<em> </em>

<em>Step 6</em>. Calculate the moles of O₂ produced

Already done in Step 4.

Moles of O₂ = 2.00 mol O₂

Option E is <em>wrong. </em>

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The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

<u>Answer:</u> The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

<u>Explanation:</u>

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                  1.8                4.7

<u>At eqllm:</u>           1.8-x             4.7-x

Evaluating the value of 'x'

\Rightarrow (1.8-x)=1.16\\\\x=0.64

The expression of K_p for above equation follows:

K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}

p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm

Putting values in above expression, we get:

K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                2.36             4.06               0.64

<u>At eqllm:</u>           2.36-x        4.06-x             0.64+x

Putting value in the equilibrium constant expression, we get:

0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

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