2CH3OH(g)→2CH4(g)+O2(g)ΔH=+252.8kJ
For a given sample of CH3OH, the enthalpy change during the reaction is 82.2kJ . How many grams of methane gas are produced?
Just treat the 82.2 kJ as a product of the reaction and use stoichiometry to found out how many moles of methane were used like so...
82.2kJ x 2mol CH4/252.8 kJ = ?
Then just multiply by the molar mass of methane (about 16.042g) to find how many grams of methane gas are produced from the reaction.
10.4 grams
Thank you!
Well, to find the number of grams of methane gas produced, we need to do a little calculation. But don't worry, I won't make this as complicated as trying to understand the stock market on a roller coaster!
Given that the enthalpy change during the reaction is +252.8 kJ, we know that this is the heat absorbed when 2 moles of CH3OH gas react. And since we're given that the enthalpy change for the given sample is 82.2 kJ, we need to find out how many moles of CH3OH reacted.
So, if we set up a little equation, we can solve for the number of moles of CH3OH:
(82.2 kJ)/(252.8 kJ) = (x moles)/(2 moles)
Now, solving for x, we find that x ≈ 0.272 moles.
Since the reaction equation tells us that 2 moles of methane (CH4) are produced for every 2 moles of CH3OH, this means that for 0.272 moles of CH3OH reacted, we'll get 0.272 moles of CH4.
But we're not done yet, my friend! We still need to find the mass of CH4. And for that, we need to use the molar mass of CH4, which is about 16.04 g/mol.
Multiply 0.272 moles by 16.04 g/mol, and you'll get approximately 4.36 grams of methane gas produced.
Voila! So, the answer is about 4.36 grams of methane gas. Keep those giggles going, my friend!
To determine the grams of methane gas produced during the reaction, we need to use the given enthalpy change and the balanced equation.
First, let's examine the balanced equation:
2CH3OH(g) → 2CH4(g) + O2(g)
From the equation, we can see that 2 moles of CH3OH produce 2 moles of CH4. This means that the mole ratio between CH3OH and CH4 is 1:1.
Next, we'll calculate the number of moles of CH4 produced. To do this, we'll use the enthalpy change given and the molar ratio derived from the balanced equation.
Given: ΔH = +252.8 kJ
Given: ΔH = 82.2 kJ
By comparing the given enthalpy change with the balanced equation, we can say that the enthalpy change of 82.2 kJ corresponds to the production of 2 moles of CH4.
Now, we need to use the molar ratio to find the number of moles of CH4 produced.
(82.2 kJ / 252.8 kJ) * 2 moles = 0.6584 moles
So, 0.6584 moles of CH4 are produced.
Finally, to find the grams of CH4, we'll use the molar mass of CH4, which is 16.04 g/mol.
Grams of CH4 = 0.6584 moles * 16.04 g/mol = 10.55 g (rounded to two decimal places)
Therefore, approximately 10.55 grams of methane gas are produced during the reaction.