Home
Class 10
CHEMISTRY
Al((s)) + fe(2)O(3(s)) to Al(2) O(3(s)) ...

`Al_((s)) + fe_(2)O_(3(s)) to Al_(2) O_(3(s)) + Fe_((s))`
(atomic masses of Al = 27U, Fe = 56U, and O = 16U)

`2Al_((s)) +Fe_(2)O_(3(s)) to Al_(2)O_(3(s)) + 2Fe_((s))` , is a balanced equation.
`(2 xx 27)U + (2xx 56 + 3xx 16)U to (2xx 27 + 3 xx 16) U + (2 xx 56) U`
`{:(54U+160U,,to102U+112U),(or2" mol" + "1 mol",,to 1"mol" + 2"mol"),(54g + 160g ,, to 102 g + 112 g):}`
Suppose that your asked to calculate the amount of aluminium, required to get 1120 kg of iron by the above reaction.

Text Solution

Verified by Experts

As per the balanced equation
Aluminium `rarr` Iron
54 g `rarr` 112 g
xg `rarr (1120 xx 1000)g`
`:. x g=((1120xx1000)g xx 54 g)/(112 g)`
`= 10000 xx 54 g`
= 540000 g or 540 kg
`:.` to get 1120 kg of iron we have to use 540 kg of aluminium.
Promotional Banner

Similar Questions

Explore conceptually related problems

Al_(s) + Fe_2O_3(s) rarr Al_2O_3(s) + Fe_(s) (atomic masses of Al = 27 U Fe = 56 U and O = 16 U).Suppose that you are asked to calculate the amount of aluminium, required to get 1120 kg of iron by the above equation.

Fe_(2)O_(3(s))+2Al_((s))rarr 2Fe_((s))+Al_(2)O_(3(s)) is a/an …………….

Molecular mass of HNO_3 (Atomic masses of H=1U, N=14U, O=16U)-

A mixture of Al_(2)O_(2) and Fe_(2) can be separated by using

A mixture of Al_(2)O_(3) and Fe_(2)O_(3) can be separated by using

Al_((s))+Fe_(2)O_(3(s))rarrAl_(2)O_(3(s))+Fe_((s)) 54_((g))+160_(g)rarr 102g+112 was given, Then, the amount of aluminium required to get 1120 grams of iron is

A mixture of Al_(2) O_(3) and Fe_(2) O_(3) can be separa-ted by using

Fe_(2)O_(3)+2AlrarrAl_(2)O_(3)+2Fe The above reaction is an example of :

Al_(2)O_(3) to AlN to A(OH)_(3) to Al_(2)O_(3) . The sequence of these products involved in

Al_(2)O_(3) to AIN to Al(OH)_(2) to Al_(2)O_(3) . The sequence of these products involved in