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devlian [24]
3 years ago
15

Blast furnaces extract pure iron from the iron(III) oxide in iron ore in a two step sequence. In the first step, carbon and oxyg

en react to form carbon monoxide: (s)(g)(g) In the second step, iron(III) oxide and carbon monoxide react to form iron and carbon dioxide: (s)(g)(s)(g) Write the net chemical equation for the production of iron from carbon, oxygen and iron(III) oxide. Be sure your equation is balanced.
Chemistry
1 answer:
zavuch27 [327]3 years ago
6 0

Answer:

6 C(s) +  3 O₂(g) + 2 Fe₂O₃(s) →  4 Fe(s) + 6 CO₂(g)

Explanation:

Iron can be formed in two steps.

Step 1: 2 C(s) + O₂(g) → 2 CO(g)

Step 2: Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g)

In order to get the net chemical equation, we will multiply the first step by 3, the second step by 2, and then add them.

6 C(s) +  3 O₂(g) → 6 CO(g)

+

2 Fe₂O₃(s) + 6 CO(g) → 4 Fe(s) + 6 CO₂(g)

--------------------------------------------------------------------------------------------------

6 C(s) +  3 O₂(g) + 2 Fe₂O₃(s) + 6 CO(g) → 6 CO(g) + 4 Fe(s) + 6 CO₂(g)

6 C(s) +  3 O₂(g) + 2 Fe₂O₃(s) →  4 Fe(s) + 6 CO₂(g)

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3 years ago
If 100. ml of water is added to 25 ml of 5.0 m nacl, the final concentration is __________.
Iteru [2.4K]
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Solid aluminum (AI) and oxygen (O_2) gas react to form solid aluminum oxide (Al_2O_3). Suppose you have 7.0 mol of Al and 9.0 mo
Nimfa-mama [501]

Answer:

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

Explanation:

Step 1: Data given

Numbers of Al = 7.0 mol

Numbers of mol O2 = O2

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Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

4Al(s) + 3O2(g) → 2Al2O3(s)  

Step 3: Calculate limiting reactant

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

Al is the limiting reactant, it will be consumed completely (7 moles).

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There will remain 9-5.25 = 3.75 moles

Step 4: Calculate moles Al2O3

For 4 moles Al we'll have 2moles Al2O3

For 7.0 moles of Al we'll have 3.5 moles of Al2O3 produced

Step 5: Calculate mass of Al2O3

Mass Al2O3 = moles Al2O3 * molar mass Al2O3

Mass Al2O3 = 3.5 moles* 101.96 g/mol

Mass Al2O3 = 356.9 grams

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

3 0
3 years ago
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× (247.80 g Ag₂S/1 mol Ag₂S) = 12.1 g Ag₂S

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3 years ago
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