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3 years ago

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Calculate the amounts requested if 3.30 mol Fe2O3 completely react according to the following equation. Fe2O3 + 2Al --> 2Fe +

Al2O3 <br />
a. moles of aluminum needed<br />
b. moles of iron formed <br />
c. moles of aluminum oxide formed

1 answer:

Natalija [7]3 years ago

4 0

The answer would be : B

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3 years ago

Convert 150 kPa to atmospheres. *

lakkis [162]

Answer: 1.48 atmosphere

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Pressure in kilopascal = 150

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3 years ago

How many grams of oxygen are required to burn 13.5 g of acetylene

Vinil7 [7]

Answer:

41.54 grams of oxygen are required to burn 13.5 g of acetylene

Explanation:

The balanced reaction is:

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

C₂H₂: 2 moles
O₂: 5 moles
CO₂: 4 moles
H₂O: 2 moles

Being the molar mass of the compounds:

C₂H₂: 26 g/mole
O₂: 32 g/mole
CO₂: 44 g/mole
H₂O: 18 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

C₂H₂: 2 moles* 26 g/mole= 52 grams
O₂: 5 moles* 32 g/mole= 160 grams
CO₂: 4 moles* 44 g/mole= 176 grams
H₂O: 2 moles* 18 g/mole= 36 grams

You can apply the following rule of three: if by stoichiometry 52 grams of acetylene react with 160 grams of oxygen, 13.5 grams of acetylene react with how much mass of oxygen?

$mass of oxygen=\frac{13.5 grams of acetylene*160 grams of oxygen}{52 grams of acetylene}$

mass of oxygen= 41.54 grams

<u><em>41.54 grams of oxygen are required to burn 13.5 g of acetylene</em></u>

<u><em></em></u>

8 0

3 years ago