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Ludmilka [50]
2 years ago
15

Part B When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 22.8 g of carbon were burned in the pres

ence of 73.8 g of oxygen, 13.0 g of oxygen remained unreacted. What mass of carbon dioxide was produced? Express your answer to one decimal place and include the appropriate units. View Available Hint(s)
Chemistry
1 answer:
aksik [14]2 years ago
6 0

<u>Answer:</u> The amount of carbon dioxide gas produced in the reaction is 83.6 grams

<u>Explanation:</u>

As, some amount of oxygen gas is left after the reaction is completed. So, it is present in excess and is considered as an excess reagent.

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

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of carbon = 22.8 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon}=\frac{22.8g}{12g/mol}=1.9mol

The chemical equation for the reaction of carbon and oxygen gas follows:

C+O_2\rightarrow CO_2

By Stoichiometry of the reaction:

1 mole of carbon produces 1 mole of carbon dioxide gas

So, 1.9 moles of carbon will produce = \frac{1}{1}\times 1.9=1.9moles of carbon dioxide gas

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 1.9 moles

Putting values in equation 1, we get:

1.9mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(1.9mol\times 44g/mol)=83.6g

Hence, the amount of carbon dioxide gas produced in the reaction is 83.6 grams

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Both 1,2−dihydronaphthalene and 1,4−dihydronaphthalene may be selectively hydrogenated to 1,2,3,4−tetrahydronaphthalene. One of
pochemuha

Answer:

1,4-dihydro = 113 kJ·mol⁻¹

1,2-dihydro = 101 kJ·mol⁻¹

Explanation:

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In 1,2-dihydronaphthalene, the 3,4-double bond is conjugated with the benzene ring.

Thus, 1,2-dihydronaphthalene is partially stabilized by resonance interactions between the ring and the double bond (think, styrene).

1,2-Dihydronaphthalene is at a lower energy level because of this stabilization.

The heat of hydrogenation of 1,2-dihydronaphthalene is therefore less than that of the 1,4-isomer when each is hydrogenated to the common product, 1,2,3,4-tetrahydronaphthalene.

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Granite is an intrusive igneous rock with large crystals because it cools slowly. Where was this rock most likely formed?
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3 years ago
Ammonia gas will react with oxygen gas to yield nitrogen monoxide gas and water vapor.
ruslelena [56]

Answer:

a) <u>0.168 moles O2</u>

<u>b) </u> <u>9.50 grams O2</u>

<u>c) 0.01662 kg NO</u>

<u>d)</u>88.9 %

Explanation:

Step 1: Data given

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation:

4NH3 + 5O2 → 4NO + 6H2O

a. How many moles of ammonia will react with 6.73g of oxygen?

Calculate moles of oxygen = mass O2/ molar mass O2

moles oxygen =  6.73 grams / 32.00 g/mol = 0.210 moles

Calculate moles of NH3

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.210 moles O2 we need 4/5 *0.210 = <u>0.168 moles O2</u>

<u />

b. If 6.42g of water is produced, how many grams of oxygen gas reacted?

Calculate moles of H2O = 6.42 grams / 18.02 g/mol = 0.356 moles

Calculate moles of O2:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.356 moles H2O we'll need 5/6 * 0.356 = 0.297 moles O2

Calculate mass of O2 = moles O2 * molar mass O2

Mass O2 = 0.297 moles O2 * 32.00 g/mol =  <u>9.50 grams O2</u>

c. If the reaction uses up 9.43105 g of ammonia, how many kilograms of nitrogen monoxide will be formed?

Calculate moles of ammonia = 9.43105 grams / 17.03 g/mol =0.5538 moles

Calculate moles of NO:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.5538 moles of NH3 we'll have 0.5538 moles NO

Calculate mass of NO

Mass NO = 0.5538 moles * 30.01 g/mol = 16.62 grams = <u>0.01662 kg NO</u>

<u />

<u />

d. When 2.51 g of ammonia react with 3.76 g of oxygen, 2.27 g of water vapor are produced. What is the percentage yield of water?

<em>Calculate moles of NH3</em> = 2.51 grams / 17.03 g/mol = 0.147 moles

<em>Calculate moles of O2 </em>= 3.76 grams / 32 g/mol = 0.118 moles

<em>Determine the limiting reactant</em>

O2 is the limiting reactant, it will completely be consumed (0.118 moles)

NH3 is in excess. There will react 4/5 * 0.118 = 0.0944 moles

There will remain 0.147 - 0.0944 = 0.0526 moles

<em>Calculate moles H2O</em>: For 0.118 moles O2 we'll have 6/5 * 0.118 = 0.1416 moles H2O

<em>Calculate mass H2O</em> = 0.1416 moles * 18.02 g/mol = 2.552 grams H2O

<em>Calculate % yield</em> = (2.27/2.552)*100 % = <u>88.9 %</u>

4 0
2 years ago
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