what volume will it occupied 40 degrees celsius a gas sample was collected when a temperature is 27 degrees celsius and the volu
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The heat of reaction (i.e. combustion) of butane (
) when reacted with oxygen (
) is -2658 kJ/mol butane, and the chemical reaction is given by:
+ 
---> 
+ 
The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants.
Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required.
Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane
Mass of butane = 0.5643 moles butane * 58 grams/mol butane
Mass of butane = 32.73 grams butane
The mass of carbon dioxide (
) can be determined by multiplying the moles of butane () with the mole ratio of () produced to the () reacted, and then with the molar mass of (), which is 44 grams/mole.
Mass of carbon dioxide produced
= 0.5643 moles butane * [4 moles/ 1 mole ] * 44 grams/mole
Mass of carbon dioxide produced
= 99.32 grams
Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams.
The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants.
Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required.
Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane
Mass of butane = 0.5643 moles butane * 58 grams/mol butane
Mass of butane = 32.73 grams butane
The mass of carbon dioxide (
Mass of carbon dioxide produced
= 0.5643 moles butane * [4 moles
Mass of carbon dioxide produced
= 99.32 grams
Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams.
The compound is formed only by sharing of electrons between the atoms. The structure of the compound is shown in the image.
Each line between two atoms represents the sharing of an electron pair which results in the formation of a single bond. Since, carbon has 4 electrons in its valence shell and hydrogen has 1 electron in its valence shell so in order to complete the octet ( to have 8 electrons in their valence shell, noble gas configuration) to attain stability carbon needs 4 more electrons and hydrogen needs 1 electron. So, sharing of electron will occur as shown in the image and the formed compound is stable in nature.
Since, the bond that is formed by sharing of electrons between atoms is known as covalent bond. So, covalent bonding is most important in .
Answer : The correct answer for the Theoretical Yield is 48.93 g of product .
Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .
Following are the steps to find theoretical yield .
Step 1) : Write a balanced reaction between Al and Cl₂ .
2 Al + 3 Cl₂→ 2 AlCl₃
Step 2: To find amount of product (AlCl₃) formed by Al .
Following are the sub steps to calculate amount of AlCl₃ formed :
a) To calculate mole of Al :
Given : Mass of Al = 35.5 g
Mole can be calculate by following formula :
Mole = 1.32 mol
b) To find mole ratio of AlCl₃ : Al
Mole ratio is calculated from balanced reaction .
Mole of Al in balanced reaction = 2
Mole of AlCl₃ in balanced reaction = 2.
Hence mole ratio of AlC; l₃ : Al = 2:2
c) To find mole of AlCl₃ formed :
Mole of AlCl₃ = 1.32 mol
d) To find mass of AlCl₃
Molar mass of AlCl₃ =
Mass of AlCl3 can be calculated using mole formula as:
Multiplying both side by
Mass of AlCl₃ = 176.00 g
Hence mass of AlCl₃ produced by Al is 176.00 g
Step 3) To find mass of product (AlCl₃) formed by Cl₂ :
Same steps will be followed to calculate mass of AlCl₃
a) Find mole of Cl₂
Mole of Cl₂ = 0.55 mol
b) Mole ratio of Cl₂ : AlCl₃
Mole of Cl₂ in balanced reaction = 3
Mole of AlCl₃ in balanced reaction = 2
Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3
c) To find mole of AlCl₃
Mole of AlCl3 = 0.367 mol
d) To find mass of AlCl₃ :
Multiplying both side by
Mass of AlCl₃ = 48.93 g
Hence mass of AlCl₃ produced by Cl₂ = 48.93 g
Step 4) To identify limiting reagent and theoretical yield :
Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .
The product AlCl₃ formed by Al = 176.00 g
The product AlCl₃ formed by Cl₂ = 48.93 g
Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .