This is stoichiometry question which involves unit conversions.
Given values:
CaO mass = 11.2g
NH4Cl mass = 22.4g
First convert the two given values to moles.
(11.2g CaO)(1mol CaO / 56.08g CaO) = 0.20mol CaO
(22.4g NH4Cl)(1mol NH4Cl / 53.49g NH4Cl) = 0.42mol NH4Cl
CaO reacts with NH4Cl on a 1:2 ratio. This means that the reactants will react in this ratio: 0.20mol CaO: 0.40mol NH4Cl. There will be an excess of 0.02mol NH4Cl unreacted because CaO is the limiting reagent (CaO is used up completely in the reaction and excess NH4Cl remains unreacted).
Now given the equation:
CaO(s) + 2 NH4Cl(s) = 2 NH3(g) + H2O(g) + CaCl2(s)
It is seen that every 2mol of NH4Cl reacts to form 2mol of NH3 because all the elements in the composition of NH3 is found in NH4Cl. This means that NH4Cl and NH3 are on a 1:1 reaction ratio. Now we use this relationship:
0.40mol NH4Cl reacted : 0.40mol NH3 produced. Convert NH3 produced to grams to find solution:
(0.40mol NH3)(17.03g NH3 / 1mol NH3) = 6.8g NH3 produced a)
We determined the excess reactant as 0.02mol NH4Cl in a). Now we convert it to grams:
(0.02mol NH4Cl)(53.49g NH4Cl / 1mol NH4Cl) = 1.07g NH4Cl unreacted b)
Answer:
The final volume will be 24.7 cm³
Explanation:
<u>Step 1:</u> Data given:
Initial temperature = 180 °C
initial volume = 13 cm³ = 13 mL
The mixture is heated to a fina,l temperature of 587 °C
Pressure and amount = constant
<u>Step 2: </u>Calculate final volume
V1/T1 = V2/T2
with V1 = the initial volume V1 = 13 mL = 13*10^-3
with T1 = the initial temperature = 180 °C = 453 Kelvin
with V2 = the final volume = TO BE DETERMINED
with T2 = the final temperature = 587 °C = 860 Kelvin
V2 = (V1*T2)/T1
V2 = (13 mL *860 Kelvin) /453 Kelvin
V2 = 24.68 mL = 24.7 cm³
The final volume will be 24.7 cm³
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desertification, urbanization, and climate change(more carbon dioxide).
Answer:
Explanation:
From the given information:
The equation for the reaction can be represented as:
The I.C.E table can be represented as:
2SO₂ O₂ 2SO₃
Initial: 14 2.6 0
Change: -2x -x +2x
Equilibrium: 14 - 2x 2.6 - x 2x
However, Since the amount of sulfur trioxide gas to be 1.6 mol.
SO₃ = 2x,
then x = 1.6/2
x = 0.8 mol
For 2SO₂; we have 14 - 2x
= 14 - 2(0.8)
= 14 - 1.6
= 12.4 mol
For O₂; we have 2.6 - x
= 2.6 - 1.6
= 1.0 mol
Thus;
[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,
[O₂] = 1/50 = 0.02 M ,
[SO₃] = 1.6/50 = 0.032 M
Kc = [SO₃]² / [SO₂]² [O₂]
= ( 0.032²) / ( 0.248² x 0.02)
= 0.8325
Recall that; the equilibrium constant for the reaction = 0.8325;
If we want to find:
Then:
Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.
