Solid iron (III) oxide reacts with hydrogen gas to form iron and water. How many grams of iron are produced when 440.23 grams of
1 answer:
When 440.23 grams of iron(III) oxide are reacted with hydrogen gas, the amount of iron produced will be 307.66 grams
<h3>Stoichiometric calculation</h3>From the equation of the reaction:
The mole ratio of iron(III) oxide to produced iron is 1:2.
Mole of 440.23 iron(III) oxide = 440.23/159.69 = 2.76 moles
Equivalent mole of produced iron = 2.76 x 2 = 5.52 moles
Mass of 5.52 moles of iron = 5.52 x 55.8 = 307.66 grams
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Answer:
Molar concentration of S₂ is 1.77×10⁻⁶M
Explanation:
For the reaction:
2H₂S(g) ⇄ 2H₂(g) + S₂(g)
The equilibirum constant, K, is defined as:
<em>(1)</em>
Concentrations in equilibirum are:
[H₂S] : 0,163/0.500L - X
[H₂] : 0,0500/0.500L + X
[S₂] : X
Replacing the concentrations and the equilibrium value in (1):
1.67x10⁻⁷ = X (X² + 0.2X + 0.01) / (X² -0.652X + 0.106)
1.67x10⁻⁷X² - 1.09x10⁻⁷X + 1.77x10⁻⁸ = X³ + 0.2X² + 0.01X
0 = X³ + 0.2X² + 0.01X - 1.77x10⁻⁸
Solving for X:
X = 1.77×10⁻⁶
As [S₂] = X, <em>molar concentration of S₂ is 1.77×10⁻⁶M</em>
I hope it helps!