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MrRa [10]
3 years ago
13

In the manufacture of steel, pure oxygen is blown through molten iron to remove some of the carbon impurity. if the combustion o

f carbon is efficient, carbon dioxide (density = 1.80 g/l) is produced. incomplete combustion produces the poisonous gas carbon monoxide (density - 1.15 g/l) and should be avoided. if you measure a gas density of 1.77 g/l, what can you conclude?
Chemistry
2 answers:
Schach [20]3 years ago
8 0

In the extraction process of steel, one of the step is purification of the iron used to make the steel. In which pure oxygen is blown on the steel at high temperature so that the carbon percentage present in the steel can be thrown out in the form of gas. The process occurs at high temperature which is called combustion process. The reaction occurs can be shown as- C(s)+O_{2}→CO_{2} (g) + CO (g). In presence of excess oxygen, the produced carbon mono oxide (CO) converts to carbon di-oxide. The reaction is CO(g) + O_{2}(g) → CO_{2} (g). From the density of the evolved gas one could identify the gas. If the gas density is 1.77g/L which is very close to the standard density of CO_{2} i.e. 1.80g/L, the gas is carbon dioxide only.    

Oksi-84 [34.3K]3 years ago
4 0

Answer:

We can conclude that the gas evolved was carbon dioxide.

Explanation:

On an efficient combustion that is on complete combustion of carbon, carbon dioxide is produced.

C+O_2\rightarrow CO_2 (Complete combustion)

But during incomplete combustion of carbon results in formation of carbon monoxide (poisons gas).

2C+O_2\rightarrow 2CO (Incomplete combustion)

Theoretical value density of carbon dioxide = 1.80 g/L

Theoretical value density of carbon monoxide = 1.15 g/L

Experimental measured density of the gas = 1.77 g/L

1.77 g/L ≈ 1.80 g/L

Since, the experimental measured density of the gas is more closer to theoretical value of density of carbon dioxide from which we can conclude that  'the gas evolved after the combustion of carbon impurity was carbon dioxide'.

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c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

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