<u>Given:</u>
Change in internal energy = ΔU = -5084.1 kJ
Change in enthalpy = ΔH = -5074.3 kJ
<u>To determine:</u>
The work done, W
<u>Explanation:</u>
Based on the first law of thermodynamics,
ΔH = ΔU + PΔV
the work done by a gas is given as:
W = -PΔV
Therefore:
ΔH = ΔU - W
W = ΔU-ΔH = -5084.1 -(-5074.3) = -9.8 kJ
Ans: Work done is -9.8 kJ
Molar mass NO₂ = 46.0 g/mol
1 mole -------- 46.0 g
2.0 moles ----- ?
Mass (NO₂) = 2.0 x 46.0 / 1
=> 92.0 g
hope this helps!
Enthalpy of formation is calculated by subtracting the total enthalpy of formation of the reactants from those of the products. This is called the HESS' LAW.
ΔHrxn = ΔH(products) - ΔH(reactants)
Since the enthalpies are not listed in this item, from reliable sources, the obtained enthalpies of formation are written below.
ΔH(C2H5OH) = -276 kJ/mol
ΔH(O2) = 0 (because O2 is a pure substance)
ΔH(CO2) = -393.5 kJ/mol
ΔH(H2O) = -285.5 kJ/mol
Using the equation above,
ΔHrxn = (2)(-393.5 kJ/mol) + (3)(-285.5 kJ/mol) - (-276 kJ/mol)
ΔHrxn = -1367.5 kJ/mol
<em>Answer: -1367.5 kJ/mol</em>
10 gm of Fe will consumes 19 gm Cl₂ and will produces 29 gm FeCl₃.
What ois Theoretical yield ?
The quantity of a product obtained from a reaction is expressed in terms of the yield of the reaction.
The amount of product predicted by stoichiometry is called the theoretical yield, whereas the amount obtained actually is called the actual yield.
- As 2 moles (111.68 g) of Fe consumes 213 gm of Cl₂ to produce 2FeCl₃
Therefore ,
10 gm of Fe will consumes = 213 / 111.68 x 10 = 19 gm Cl₂
- As 2 moles (111.68 g) of Fe produces 2 mole (324 gm) of FeCl₃
Therefore ,
10 gm of Fe will produces = 324 / 111.68 x 10 = 29 gm FeCl₃
Hence , 10 gm of Fe will consumes 19 gm Cl₂ and will produces 29 gm FeCl₃.
Learn more about Theoretical yield here ;
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<span>a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.
</span>and % will be 60%.
