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Yuki888 [10]
3 years ago
11

A thermite reaction realeses large amounts of heat and light resulting in the melting of the iron metal

Chemistry
1 answer:
lara [203]3 years ago
4 0

Answer:

1. Fe₂O₃ : Al = 1 : 2

2. Fe₂O₃ : Al₂O₃ = 1 : 1

3. Fe₂O₃ : Fe = 1 : 2

Note: The question is incomplete. The complete question is given below :

A thermite reaction releases large amounts of heat and light, resulting in the melting of the iron metal that forms during the reaction. The balanced chemical equation is given below: Fe₂O₃(s) + 2Al(s) ---> Al₂O₃(s) + 2Fe(l) Determine the correct mole ratios for the following substances, based on this chemical equation.

Explanation:

A thermite reaction is a highly exothermic reaction that occurs between a mixture of powdered aluminium and iron (iii) oxide.

The reaction is a redox reaction which produces aluminium oxide and metallic iron.

The heat given out in the reaction melts the iron formed. It is used to join railway tracks or cracked machine parts. The molten iron runs down between the tracks and welds them together.

The balanced equation of the reaction is given below :

Fe₂O₃(s) + 2Al(s) ---> Al₂O₃(s) + 2Fe(l) + Heat

The mole ratio of the reactant and products are as follows:

1. 1 mole of Iron (iii) oxide and 2 moles of aluminium are combined

Fe₂O₃ : Al = 1 : 2

2. 1 mole of aluminium oxide is produced from 1 mole of iron (iii) oxide

Fe₂O₃ : Al₂O₃ = 1 : 1

3. 2 moles of of iron are produced from 1 mole of iron (iii) oxide

Fe₂O₃ : Fe = 1 : 2

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Explanation:

V=IR

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7 0
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What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa
almond37 [142]

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of stomach acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL

Putting values in above equation, we get:

1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL

Hence, the volume of HCl neutralized is 1.25 mL

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3 years ago
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R = 0.08206 L atm / mol K

n = (0.720 atm)(1.90 L) / (0.08206 L atm / mol K)(306 K) = 0.0545 mol of gas

Now divide grams by mol to get the molecular weight.

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