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11111nata11111 [884]
3 years ago
8

Gaseous methane (CH4) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H20). What

is the theoretical yield of carbon dioxide formed from the reaction of 1.28 g of methane and 10.1 g of oxygen gas? Be sure your answer has the correct number of significant digits in it. 02
Chemistry
1 answer:
Mariana [72]3 years ago
3 0

Answer:

Theoretical yield = 3.51 g

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

For CH_4  :-

Mass of CH_4  = 1.28 g

Molar mass of CH_4  = 16.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{1.28\ g}{16.04\ g/mol}

Moles_{CH_4}= 0.0798\ mol

For O_2  :-

Mass of O_2  = 10.1 g

Molar mass of O_2  = 31.998 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{10.1\ g}{31.998\ g/mol}

Moles_{O_2}= 0.3156\ mol

According to the given reaction:

CH_4+2O_2\rightarrow CO_2+2H_2O

1 mole of methane gas reacts with 2 moles of oxygen gas

0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas

Moles of oxygen gas = 0.1596 moles

Available moles of oxygen gas = 0.3156 moles

<u>Limiting reagent is the one which is present in small amount. Thus, CH_4 is limiting reagent. </u>

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction produces 1 mole of carbon dioxide.

0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.

Mole of carbon dioxide = 0.0798 mole

Molar mass of carbon dioxide = 44.01 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0798\ moles= \frac{Mass}{44.01\ g/mol}

Mass of CO_2 = 3.51 g

<u> Theoretical yield = 3.51 g</u>

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