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garik1379 [7]
3 years ago
7

Iron forms a series of compounds of the type Fex(CO)y. In air, they are oxidized to Fe2O3and CO2gas. After heating a 0.142 g sam

ple of Fex(CO)yin air, you isolate the CO2in a 1.50 L flask at 25 °C. The pressure of the gas is 44.9 mmHg. What is the empirical formula of Fex(CO)y
Chemistry
1 answer:
Elodia [21]3 years ago
6 0

Answer:

The empirical formula is = Fe(CO)_2

Explanation:

Carbon dioxide obtained:

Pressure = 44.9 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 44.9 / 760 = 0.0591 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25 + 273.15) K = 298.15 K  

V = 1.50 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.0591 atm × 1.50 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 0.0036 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.0036 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.0036 x 12.0107 = 0.0432 g

Given that the compound only contains iron and carbon. So,

Mass of Fe in the sample = Total mass - Mass of C

Mass of the sample = 0.142 g

Mass of Fe in sample = 0.142 g - 0.0432 g = 0.0988 g  

Molar mass of Fe = 55.845 g/mol

Moles of Fe  = 0.0988  / 55.845  = 0.0018 moles

Taking the simplest ratio for Fe and C as:

0.0018 : 0.0036

 = 1 : 2

The empirical formula is = Fe(CO)_2

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Na2CO3(aq) + CaCl2(aq) — 2 NaCl(aq) + CaCO3(s)
attashe74 [19]

Answer:

100 mL

Explanation:

Given data:

Mass of CaCO₃ produced = 2.00 g

Molarity of CaCl₂ = 0.200 M

Volume of CaCl₂ needed = ?

Solution:

Chemical equation:

Na₂CO₃  + CaCl₂    →      2NaCl + CaCO₃

First of all we will calculate the number of moles of CaCO₃.

Number of moles = mass/molar mass

Number of moles = 2.00 g / 100.09 g/mol

Number of moles = 0.02 mol

Now we will compare the moles of CaCO₃ and CaCl₂.

              CaCO₃          :            CaCl₂

                   1                :               1

                 0.02           :              0.02

Thus, 0.02 moles of CaCl₂ react,

Volume of CaCl₂ reacted:

Molarity = number of moles / volume in L

0.200 M = 0.02 mol / volume in L

Volume in L = 0.02 mol / 0.200 M

Volume in L = 0.1 L

Volume in mL:

0.1 L × 1000 mL/1L

100 mL

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