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iogann1982 [59]
3 years ago
12

You are working in your laboratory and decide to do some cleaning. You find a test tube with some brown substance congealed at t

he bottom. You take a piece of iron wool to scrub it out, but cannot remove it. You leave the iron wool in the test tube and set it aside to clean your beakers instead. You do not realize the test tube is next to the heater and after a few minutes, a brown gas starts smoking out of the test tube, followed by a reddish-brown gas. Upon observation, the iron wool is coated in a yellow-brown substance. You realize that the mystery substance was _________.
A. Argon (Ar)
B. Scandium (Sc)
C. Bromine (Br)
D. Iodide (I)
Chemistry
1 answer:
Anna [14]3 years ago
7 0

Answer:

  • <em>The mystery substance is</em> <u>C. Bromine (Br) </u>

Explanation:

<em>Argon (Ar) </em>is a noble gas. Whose freezing point is -189 °C (very low), thus it cannot be the frozen substance. Also, it is not reactive, thus is would have not reacted with iron. Hence, argon is not the mystery substance.

<em>Scandium (Sc) </em>is a metal from group 3 of the periodic table, thus is will not react with iron. Thus, scandium is not the mystery substance.

Both <em>bromine</em> and <em>iodine</em> are halogens (group 17 of the periodic table).

The freezing point of bromine is −7.2 °C, ​and the freezing point of iodine is 113.7 °C. Thus, both could be solids (frozen) in the lab.

The reactivity of the halogens decrease from top to bottom inside the group. Bromine is above iodine. Then bromine is more reactive than iodine.

Bromine is reactive enough to react with iron. Iodine is not reactive enough to react with iron.

You can find in the internet that bromine vapour over hot iron reacts  producing iron(III) bromide. Also, that bromine vapors are red-brown.

Therefore, <em>the mystery substance is bromine (Br).</em>

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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

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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

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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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