The balanced chemical equation that illustrates this reaction is:
<span>C2H4 + 3O2 --> 2CO2 + 2H2O
</span>
From the periodic table:
mass of carbon = 12 grams
mass of hydrogen = 1 gram
Therefore:
molar mass of C2H4 = 12(2) + 4(1) = 24 + 4 = 28 grams
number of moles = mass / molar mass
number of moles of C2H4 = 54.7 / 28 = 1.95 moles
From the balanced equation above:
3 moles of oxygen are required to react with one mole of C2H4, therefore, to know the number of moles required to react with 1.95 moles of C2H4, all you have to do is cross multiplication as follows:
number of oxygen moles = (1.95*3) / 1 = 5.85 moles
The grams of ethane present in a sample containing 0.4271 mole is 12.84 g
<h3>Description of mole </h3>
The mole of a substance is related to it's mass and molar mass according to the following equation
Mole = mass / molar mass
With the above formula, we can obtain the mass of ethane. Details below
<h3>How to determine the mass of ethane</h3>
The following data were obtained from the question:
- Mole of ethane = 0.4271 mole
- Molar mass of ethane = 30.067 g/mol
- Mass of ethane =?
The mass of ethane present in the sample can be obtained as follow:
Mole = mass / molar mass
Cross multiply
Mass = mole × molar mass
Mass of ethane = 0.4271 × 30.067
Mass of ethane = 12.84 g
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Your answer would be C9H16
Answer:
Calcium (Ca) => 2
Aluminium (Al) => 3
Rubidium (Rb) => 2
Oxygen (O) => 2
Sulphur (S) => 2
Iodine (I) => 1
<em><u>Formulae</u></em>
<em>Calcium</em><em> </em><em>oxide</em><em> </em><em>:</em><em> </em>CaO
<em>Alumin</em><em>ium</em><em> </em><em>iodide</em><em> </em><em>:</em><em> </em>AlI3
<em>Rubidi</em><em>um</em><em> </em><em>sulphide</em><em> </em><em>:</em><em> </em>RbS
<em>Alum</em><em>inium</em><em> </em><em>oxide</em><em> </em><em>:</em><em> </em>Al2O3
