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kvv77 [185]
3 years ago
8

To find the formula of a compound composed of iron and carbon monoxide, Fex(CO)y, the compound is burned in pure oxygen, an reac

tion that proceeds according to the following unbalanced equation.
Fex(CO)y + O2 --> Fe2O3 + CO2

If you burn 1.959 g. of Fex(CO)y and obtain 0.799 g. of Fe2O3 and 2.200 g. of CO2, what is the empirical formula of Fex(CO)y?
Chemistry
1 answer:
Anarel [89]3 years ago
8 0

Answer:

The empirical formula is: Fe(CO)₅

Explanation:

According the global reaction:

Feₓ(CO)y + O₂ → Fe₂O₃ + CO₂

You should calculate Fe₂O₃ and CO₂ moles, thus:

0,799 Fe₂O₃ grams  × \frac{1 mole}{159.69 Fe2O3 g} = 5,00×10⁻³ Fe₂O₃ moles

2,200 CO₂ grams  × \frac{1 mole}{44,01 CO2 g} = 5,00×10⁻²CO₂ moles

The ratio between Fe₂O₃ moles and CO₂ moles is 1:10. Thus ratio between x and y must be 1:5 because Fe₂O₃ has 2 irons but CO₂ has just one carbon.

Assuming the formula is Fe₁(CO)₅ the molecular weight is 195,9 g/mol. Thus:

1,959 Fe(CO)₅ grams  × \frac{1 mole}{195,9 Fe(CO)5 g} = 1,00×10⁻² Fe(CO)₅ moles

Thus, assuming 1,00×10⁻² moles as basis for calculation, the global reaction is:

1 Fe(CO)₅ + ¹³/₂O₂ → ¹/₂ Fe₂O₃ + 5 CO₂

With this balanced equation the moles produced have sense, thus, the empirical formula is: Fe(CO)₅

I hope it helps!

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A gas has a volume of 5.00 L at 0°C. What final temperature, in degrees Celsius, is needed to change the volume of the gas to ea
Diano4ka-milaya [45]

Answer:

A = -213.09°C

B = 15014.85 °C

C = -268.37°C

Explanation:

Given data:

Initial volume of gas = 5.00 L

Initial temperature = 0°C  (273 K)

Final volume = 1100 mL, 280 L, 87.5 mL

Final temperature = ?

Solution:

Formula:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Conversion of mL into L.

Final volume = 1100 mL/1000 = 1.1 L

Final volume =  87.5 mL/1000 = 0.0875 L

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 1.1 L × 273 K / 5.00 L

T₂ = 300.3 L.K / 5.00 K

T₂ = 60.06 K

60.06 K - 273 = -213.09°C

2)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 280 L × 273 K / 5.00 L

T₂ = 76440 L.K / 5.00 K

T₂ = 15288 K

15288 K - 273 = 15014.85 °C

3)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 0.0875 L × 273 K / 5.00 L

T₂ = 23.8875 L.K / 5.00 K

T₂ = 4.78 K

4.78 K - 273 = -268.37°C

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