Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
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NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
Answer:
Explanation:
The gas ideal law is
PV= nRT (equation 1)
Where:
P = pressure
R = gas constant
T = temperature
n= moles of substance
V = volume
Working with equation 1 we can get

The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.
or
(equation 2)
The cylindrical container has a constant pressure p
The volume is the volume of a cylinder this is

Where:
r = radius
h = height
(pi) = number pi (3.1415)
This cylinder has a radius, r and height, h so the volume is 
Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:
Replacing these values in the equation 2 we get:
(equation 2)
This may help you
<span>You need to use some stoichiometry here. The only way to do that is if you're working in moles. Since you're given grams of Al, you can convert that moles by dividing by the molar mass.
Then from looking at the coefficients in your equation, you can see that for however many moles of Al react, the same numbers of moles of Fe will be produced, but only half as many moles of Al2O3 will be produced.
To go back to grams, multiply the moles of each product that you get by their molar masses!</span>
Answer:
D
Explanation:
It would be D because you are observing the reaction and don’t change anything