The ratio of $F e_{2} O_{3}$ and $A l$ by weight, in thermite is about:

1 : 3

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1 : 2

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3 : 1

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none of the above

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Solution

The correct option is C $3 : 1$
Thermite is the mixture of $F e_{2} O_{3}$ and $A l$ . Due to great affinity of aluminium toward oxygen, it readily combines with oxygen. Hence, Goldsmith used $A l$ to reduce metal oxides in extraction. In thermite the ratio of $F e_{2} O_{3}$ and $A l$ is taken $3 : 1$ by weight.
$F e_{2} O_{3} + 2 A l \to 2 F e + A l_{2} O_{3}$
$(2 \times 56 + 3 \times 16 = 160) (2 \times 27 = 54)$

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Similar questions

Q. The ratio of

F e_{2} O_{3}

and Al, in thermite is:

Iron oxides $F e_{2} O_{3}$ can be reduced by $A l$ in the Goldschmidt thermite process.

Q. an intimate mixture of ferric oxide and aluminium is used as solid rocket fuels.calculate fuel value per gram and fuel value per cc of the mixture.heats of formation and densities are as follows;

dHf al2o3= 399 kcal/mole
dhf fe2o3=199kcal/mole density of fe2o3=5.2g/cc density of al=2.7gm/cc

soln;-
2al +3/2 o2 -->al2o3 eq 1
2fe + 3/2 o2---- fe2o3 eg 2
2al +fe2o3= al2o3 + 2fe

1*eq 1 - 1*eq 2
dh=-399-(-199)
= -200 kcal

fuel value/gram
2 moles of al + 1 mol fe2o3
2*27 + 1 * 160
214gm

; fuel value /gram = 200/214= 0.93kcal/gm
fuel gram/ cc
2 mole of al= 54/2.7= 20cc
1 mole of fe2o3 = 214/5.2=30.7cc // here ,1 mole of fe2o3 = 160 ...... and 214gm is the sumation of both al
and fe2o3 ..... while if we multiply 30.7*5.2 =159.64 which is near about 160

Q. Which of the following options represents the reducing agent and the substance that gets oxidised during a thermite reaction?

Option	Substance oxidised	Reducing agent
A	Al	Fe
B	${A l}_{2} O_{3}$	Al
C	Al	Al
D	${F e}_{2} O_{3}$	Fe

[1 mark]

${F e}_{2} O_{3}$ + $A l$ $\to$ $F e$ + ${A l}_{2} O_{3}$
This is the reaction used in thermite welding process. Calcium can also be used instead of aluminium in this process.

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The ratio of Fe2O3 and Al by weight, in thermite is about:

The ratio of $F e_{2} O_{3}$ and $A l$ by weight, in thermite is about: