Respuesta :

anfabba15

Answer:

2.66 g of Fe, can be obtained from the reaction

Explanation:

Let's think the reaction:

2Fe₂O₃   +  6CO  →   4Fe +  6CO₂

Ratio is 2:4, so If i have x moles of iron (III) oxide, I will produce the double of moles of Fe.

Mass / Molar mass = Mol

3.80 g / 159.7 g/m = 0.0237 moles

0.0237 moles . 2 = 0.0475 moles

Molar mass Fe = 55.85 g/m

Mol . Molar mass = Mass → 0.0475 m . 55.85 g/m = 2.66 grams

2 moles of Iron oxide produce, 4 moles of iron hence the molar ratio is 1:2. The mass of iron obtained from 3.08 g of iron(III) oxide is 2.66 g.  

 

The reaction:  

[tex]\bold {2Fe_2O_3 + 6CO \rightarrow 4Fe + 6CO_2}[/tex]  

2 moles of Iron oxide produce, 4 moles of iron hence the molar ratio is 1:2.

Molar mass of  Fe = 55.85 g/m

Molar mass of  iron (III) oxide = 159.7 g/mol.

Moles of iron oxide,  

[tex]\bold {\dfrac w{m} = \dfrac {3.80\ g}{159.7\ g/mole} = 0.0237\ moles }\\[/tex]  

So, moles of the iron,  

[tex]\bold {0.0237\ moles \times 2 = 0.0475\ moles}[/tex]  

Molar mass of  Fe = 55.85 g/m  

Thus, mass of iron,

[tex]\bold {w = n\times m = 55.85 g/mol \times 0.0475 m = 2.66 g}[/tex]

Therefore, the mass of iron obtained from 3.08 g of iron(III) oxide is 2.66 g.

To know more about molar mass,

https://brainly.com/question/12127540

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