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Aleksandr-060686 [28]
3 years ago
15

Consider the reaction of NO and CO to form N2 and CO2, according to the balanced equation: 2 NO (g) + 2 CO (g) → N2 (g) + 2 CO2

(g) Identify the excess reactant, the limiting reactant, and the product(s) using the molecular art. (Black spheres are carbon, blue spheres are nitrogen, and red spheres are oxygen.)
Chemistry
1 answer:
Gekata [30.6K]3 years ago
3 0

The image is not given in the question, it is attached below:

<u>Answer:</u> The excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

<u>Explanation:</u>

In the given image:

Red spheres represent oxygen atoms, blue spheres represent nitrogen atoms and black spheres represent carbon atoms

The combination of 1 black and 2 red spheres will represent carbon dioxide (CO_2) compound

The combination of 2 blue spheres will represent nitrogen molecule (N_2)

The combination of 1 blue and 1 red sphere will represent nitrogen monoxide (NO) compound

The combination of 1 black and 1 red sphere will represent nitrogen monoxide (NO) compound

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

We are given:

Given moles of NO = 6 moles

Given moles of CO = 4 moles

For the given chemical equation:

2NO(g)+2CO(g)\rightarrow N_2(g)+2CO_2(g)

By stoichiometry of the reaction:

If 2 moles of CO reacts with 2 moles of NO

So, 4 moles of CO will react with = \frac{2}{2}\times 4=4mol of NO

As the given amount of NO is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, CO is considered a limiting reagent because it limits the formation of the product.

Hence, the excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

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7.21 × 10⁴ J

Explanation:

Ethanol is solid below -114.5°c, liquid between -114.5°C and 78.4°C, and gaseous above 78.4°C.

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We need to calculate the heat required in different stages and then add them.

The moles of ethanol are:

48.3g.\frac{1mol}{46.07g} =1.05mol

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Q₁ = ΔHfus . n = (4.60 kJ/mol) . 1.05 mol = 4.83 kJ = 4.83 × 10³ J

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Liquid: from -114.5°C to 78.4°C

Q₂ = c(l) . m . ΔT = (2.45 J/g.°C) . 48.3g . [78.4°C-(-114.5°C)] = 2.28 × 10⁴ J

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Q₄ = c(g) . m . ΔT = (1.43 J/g.°C) . 48.3g . (135.3°C-78.4°C) = 3.93 × 10³ J

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Cl                          3.0                       Cl-F                      1.0

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F                            4.0                       Cl-Cl                      0

H                            2.1                       H-C                       0.4

C                            2.5                       H-N                       0.9

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O                             3.5                      Br-F                       1.2

I                               2.7                      I-F                         1.3

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