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kompoz [17]
3 years ago
6

In the oxidation of iron; how many grams of iron (III) oxide will be produced from 6.20 mol of Fe?

Chemistry
1 answer:
Flura [38]3 years ago
8 0

Answer:

496 g of Fe₂O₃.

Explanation:

The balanced equation for the reaction is given below:

4Fe + 3O₂ —> 2Fe₂O₃

From the balanced equation above,

4 moles of Fe reacted to produce 2 moles of Fe₂O₃.

Therefore, 6.20 moles of Fe will react to produce = (6.20 × 2)/4 = 3.1 moles of Fe₂O₃

Finally, we shall determine the mass of 3.1 moles of Fe₂O₃. This can be obtained as follow:

Mole of Fe₂O₃ = 3.1 moles

Molar mass of Fe₂O₃ = (56 × 2) + (3×16)

= 112 + 48

= 160 g/mol

Mass of Fe₂O₃ =?

Mass = mole × molar mass

Mass of Fe₂O₃ = 3.1 × 160

Mass of Fe₂O₃ = 496 g

Therefore, 496 g of Fe₂O₃ were produced from the reaction.

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Using the equations 2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)∆H° = -6271 kJ/mol C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/m
Dafna11 [192]

Answer : The enthalpy for the reaction is 49.1 kJ/mol

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_6H_6 will be,

6C(s)+3H_2(g)\rightarrow C_6H_6(l)    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) 2C_6H_6(g)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)    

\Delta H_1=-6271kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    

\Delta H_2=-393.5kJ/mole

(3) 2H_2(g)+O_2(g)\rightarrow 2H_2O(l)    

\Delta H_3=-483.6kJ/mole

Now we will reverse the reaction 1 and divide by 2, multiply reaction 2 by 6 and reaction 3 by 3/2 then adding all the equations, we get :

(1) 6CO_2(g)+3H_2O(l)\rightarrow C_6H_6(g)+\frac{15}{2}O_2(g)    

\Delta H_1=-\frac{-6271kJ/mole}{2}=3135.5kJ/mol

(2) 6C(s)+6O_2(g)\rightarrow 6CO_2(g)    

\Delta H_2=6\times (-393.5kJ/mole)=-2361kJ/mol

(3) 3H_2(g)+\frac{3}{2}O_2(g)\rightarrow 3H_2O(l)    

\Delta H_3=\farc{3}{2}\times (-483.6kJ/mole)=-725.4kJ/mol

The expression for enthalpy of formation of C_6H_6 will be,

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(3135.5kJ/mole)+(-2361kJ/mole)+(-725.4kJ/mole)

\Delta H=49.1kJ/mole

Therefore, the enthalpy for the reaction is 49.1 kJ/mol

7 0
3 years ago
328. mL of 0.00345 M NaI (aq) is combined with 703. mL of 0.00802 M Pb(NO3)2 (aq). Determine if a precipitate will form given th
Vsevolod [243]

Answer:

The solution will not form a precipitate.

Explanation:

The Ksp of PbI₂ is:

PbI₂(s) ⇄ 2I⁻(aq) + Pb²⁺(aq)

Ksp = 1.40x10⁻⁸ = [I⁻]²[Pb²⁺] <em>Concentrations in equilibrium</em>

When 328mL of 0.00345M NaI(aq) is combined with 703mL of 0.00802M Pb(NO₃)₂. Molar concentration of I⁻ and Pb²⁺ are:

[I⁻] = 0.00345M × (328mL / (328mL+703mL) =<em> 1.098x10⁻³M</em>

[Pb²⁺] = 0.00802M × (703mL / (328mL+703mL) =<em> 5.469x10⁻³M</em>

<em />

Q = [I⁻]²[Pb²⁺] <em>Concentrations not necessary in equilibrium</em>

If Q = Ksp, the solution is saturated, Q > Ksp, the solution will form a precipitate, if Q < Ksp, the solution is not saturated.

Replacing:

Q = [1.098x10⁻³M]²[5.469x10⁻³M] = 6.59x10⁻⁹

As Q < Ksp, the solution is not saturated and <em>will not form a precipitate</em>.

6 0
3 years ago
Which of the following most likely requires intermolecular forces?
castortr0y [4]
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Answer:

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Explanation:

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