Balance Chemical Equation for this reaction is,
2 CH₄ + O₂ → 2CH₃OH
According to this eq, 22.4 L (1 moles) of Oxygen requires 44.8 L (2 mole) CH₄ for complete reaction.
So, the volume of CH₄ required to consume 0.66 L of O₂ is calculated as,
22.4 L O₂ required to consume = 44.8 L CH₄
0.660 L O₂ will require = X L of CH₄
Solving for X,
X = (44.8 L × 0.660 L) ÷ 22.4 L
X = 1.320 L of CH₄
Result:
1.320 L of CH₄ <span>gas is needed to react completely with 0.660 L of O</span>₂<span> gas to form methanol (CH</span>₃OH<span>).</span>
Answer :
- Carbonyl group : It is a functional group composed of a carbon atom that double bonded to oxygen atom. It is represented as

Carboxylic group : It is the class of organic compound in which the carboxylic (-COOH) group is attached to a hydrocarbon is known as carboxylic.
The general formula of carboxylic is,
. According to the IUPAC naming, the carboxylic are named as alkanoic acids.
Aldehyde group : It is the class of organic compound in which the (-CHO) group is attached to a hydrocarbon is known as aldehyde.
The general representation of aldehyde is,
. According to the IUPAC naming, the aldehyde are named as alkanals.
Ketone group : It is the class of organic compound in which the (-CO) group is directly attached to the two alkyl group of carbon is known as ketone.
The general representation of ketone is,
. According to the IUPAC naming, the ketone are named as alkanone.
Ester group : It is the class of organic compound in which the (-COO) group is directly attached to the two alkyl group of carbon is known as ester.
The general representation of ester is,
. According to the IUPAC naming, the ester are named as alkyl alkanoate.
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Answer:
44.8 L
Explanation:
Using the ideal gas law equation:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
At Standard temperature and pressure (STP);
P = 1 atm
T = 273K
Hence, when n = 2moles, the volume of the gas is:
Using PV = nRT
1 × V = 2 × 0.0821 × 273
V = 44.83
V = 44.8 L
MgF2 is <span>Magnesium fluoride
hope this helps!</span>