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NikAS [45]
3 years ago
13

Iron and carbon monoxide are made by heating 5.53 kg of iron ore, Fe₂O₃, and carbon. What is the theoretical yield of iron in ki

lograms?
Chemistry
1 answer:
vfiekz [6]3 years ago
3 0

Answer:

Mass of Fe produced = 3.86785211069  ≈ 3.87 kg

Explanation:

From the question the chemical reaction can be written as follow :

Fe2O3 + C → Fe + CO  

Balance the equation

Fe2O3 + 3C → 2Fe + 3CO  

compute the molecular mass of Fe2O3 and atomic mass of Iron(Fe)

Molecular mass of Fe2O3 = 55.845(2) + 15.999(3) = 111.69 + 47.997 = 159.687 g

Atomic mass of iron = 55.845 g

From the balanced equation

159.687 g of Fe2O3  produces 2 ×  55.845 = 111.69 g of Fe(iron)

Convert the 5.53 kg to gram

1 kg = 1000 g

5.53 kg  = 5.53 × 1000 = 5530 g

since,

159.687 g of Fe2O3  produces 2 ×  55.845 = 111.69 g of Iron(Fe)

5530 g of  Fe2O3  will produce  

cross multiply

Mass of Fe produced =5530 × 111.69/159.687

Mass of Fe produced = 617645.7

/159.687

Mass of Fe produced = 3867.85211069  g

convert to kg

1000 g = 1 kg

3867.85211069   = 3867.85211069/1000

Mass of Fe produced = 3.86785211069  ≈ 3.87 kg

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Explanation:

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5 0
3 years ago
When radium-226 (atomic number-88) decays by emitting an alpha particle it becomes _____.
topjm [15]

Answer:

d. Radon-222

Explanation:

²²⁶₈₈Ra → ²²²₈₆Rn  + ⁴₂He

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3 0
3 years ago
What does one mole of oxygen and one mole of carbon dioxide have in common?
arsen [322]

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Equal number of moles

Explanation:

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5 0
3 years ago
During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce
swat32

Answer: 16.32 g of O_2 as excess reagent are left.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol

\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)  

According to stoichiometry :

2 moles of SO_2 require = 1 mole of O_2

Thus 0.34 moles of SO_2 will require=\frac{1}{2}\times 0.34=0.17moles  of O_2

Thus SO_2 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

Moles of O_2 left = (0.68-0.17) mol = 0.51 mol

Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

Thus 16.32 g of O_2 as excess reagent are left.

3 0
3 years ago
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