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enter image description here

c(O) = 0.40 wt% C

c($\alpha$) = 0.022 wt% C

c(FeC3) = 6.70 wt% C3

For the Fe-C alloy with 99.6 wt% Fe-0.40 wt% C, determine (at the temperature immediately below the eutectoid) determine the amount of cementite formed (in g) per 100 g of steel

Solution:

$$\frac{Fe_3C}{Fe_3C+\alpha} =\frac{C_0 - C_{\alpha}}{C_{Fe_3C}-C_{\alpha}}.100 = \frac{0.4-0.022}{6.7-0.022}.100 = 5.7g$$

  • $Fe_3C = 5.7g$
  • $\alpha = 94.3g$

but i though that since its 100g steel we got that alpha=(0.022 * 100) and Fe3C=(6.7 * 100)g . why is this wrong?

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    $\begingroup$ +1 Nice first question!. Next time can you use MathJaX rather than copying the images directly. It would make the question searchable for other users. $\endgroup$
    – Anoop A Nair
    Jan 16, 2021 at 6:20

1 Answer 1

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I think you are considering the carbon % in alpha-solid solution and Fe3C as the overall percentages of alpha-solid solution and Fe3C in the alloy. For example - 6.7% is the carbon amount in Fe3C, it is not the overall % of Fe3C in the alloy.So you can't just multiply 6.7 with 100 and indicate it as the amount of Fe3C in the alloy.

Hope it helps.

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