In the reaction 2Al + 3 MnO ---> Al2O3 +3Mn
220g Al and 400g of MnO are mixed,
which substance is remained in access after the reaction and by how much \?
Ive got that Al will remain at the en by 4.32g
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at the end of the question after 4.32 its Moles not gram
please check my answer
Mn will remain in exces as it is 5.7mol and Al2O3 is 1.90 moles so we can say mn will be in excess by 3.8 ok
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ha and they are asking which remain not which get consume :P
how can u say that Al2O3 is 1.9 mole
cal krk :P
arre haan wo to theek hai
=> 5.6 ---> 5.6/3
look u find limiting reagent is MNO ok now u see its coefficeint is 3 ant its mole is 5.7 now cal that AL2O3 coefficeint is 1 so its mole is 1.90 now u see for Mn in a same way and u find coeffiecint is same dat is 3 means its mole is 5.7 now u see that no. of mole of Mn is max dan Al2O3
whose mole max means excess amount :) and by how much jusst subtract the value :)
now for 1 Al2O3 -- 2 Al
==> 1.9 Al2O3-- 3.8 Al
now left is 8.14-3.8= 4.32
ye mera DIMAAAG hai
arrey limiting reaent is MnO so y u touch Al
maine question hi galat samajh liya tha
ek baat batao Al kitna use hua
main sirf reactants dekh rha th jabki last me product bhi to hai
dekho Al k mole 8.4 h and MnO k 5.7 now u see MnO is limiting reagent here means 5.7 mole i used from both of the reactantss ok
u knw VJ sir k chem he is awesome** :)