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Lady bird [3.3K]
3 years ago
11

The emission of NO₂ by fossil fuel combustion can be prevented by injecting gaseous urea into the combustion mixture. The urea r

educes NO (which oxidizes in air to form NO₂) according to the following reaction: 2CO(NH₂)₂(g)+4NO(g)+O₂(g)→4N₂(g)+2CO₂(g)+4H₂O(g) Suppose that the exhaust stream of an automobile has a flow rate of 2.37 L/s at 657 K and contains a partial pressure of NO of 14.0 torr .What total mass of urea is necessary to react completely with the NO formed during 8.1 hours of driving?Express answer in g.
Chemistry
1 answer:
liq [111]3 years ago
4 0

Answer:

There is 709.1 grams of urea needed.

Explanation:

<u>Step 1:</u> The balanced equation:

2CO(NH2)2(g)+4NO(g)+O2(g)→4N2(g)+2CO2(g)+4H2O(g)

<u>Step 2:</u> Data given

The exhaust stream of an automobile has a flow rate of 2.37 L/s

This happens at a temperature of 657 Kelvin

The partial pressure of NO is 14.0 torr

<u>Step 3:</u> Calculating the volume during 8.1 hours

V = 2.37L/s ∙ 8.1hr ∙3600s/hr = 69109.2L = 69.1092m³

<u>Step 4</u>: Calculate  the partial pressure of nitric oxide:

p = 14 torr ∙ 101325Pa/760torr = 1866.5Pa

<u>Step 5:</u> Calculating number of moles of NO

⇒ We use the ideal gas law P*V=n*R*T

n(NO) = P*V/R*T

with P= The partial pressure of NO = 1866.5 Pa

with V =the volume of NO = 69.1092m³

with R = the gas constant = 8.314472 Pa*m³/mol*K

with T = 657 Kelvin

n(NO) = 1866.5Pa ∙ 69.1092m³ / (8.314472Pa*m³/mol*K ∙ 657K)

= 23.61 mol es

<u>Step 6:</u> Calculate moles of urea

Since there is consumed 2 moles of urea per 4 moles of nitric oxide.

This means for 24.432 moles of NO, there is consumed 23.61 /2 = 11.806 moles of urea.

<u>Step 7</u>: Calculate mass of urea

m(CO(NH₂)₂) = 11.806 moles ∙ 60.06g/moles = 709.07g

There is 709.1 grams of urea needed.

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52.09×5/18

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Answer:

The answer to your question is 92.7%

Explanation:

Balanced Chemical reaction

                             3 Zn  + Fe₂(SO₄)₃   ⇒   2Fe   +   3ZnSO₄

Molecular weight

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                           196.2 g of Zinc ------------------ 112 g of Iron

                            20.4 g of Zinc  -----------------   x

                            x = (20.4 x 112) / 196.2

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% yield = \frac{10.8}{11.65}  x 100

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<h3>Explanation:</h3>
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Let's replace → 0.899°C = 0.511 °C/m . m . 2

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