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Lostsunrise [7]
2 years ago
7

Help can someone do this for me

Chemistry
2 answers:
Pie2 years ago
7 0

Answer:

Magnesium Mg 12  12

Phosphorous P 15  15

Fluorine F 9  9

Iron Fe 26  26

Calcium Ca 20  20

Fluorine F  9 9

denis23 [38]2 years ago
4 0

Answer:

Magnesium Mg 12  12

Phosphorous P 15  15

Flourine F 9  9

Iron Fe 26  26

Calcium Ca 20  20

Flourine F  9 9

Explanation:

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Explanation:

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2 years ago
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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

Molar mass of Al = 26.98 g/mol

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).

O2 is in excess.  There will react 3/4 * 7 = 5.25 moles

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)

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of course migration

Explanation:

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C - specific heat.

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