Question

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 end by 4.32 moles

Answer 1

2 moles of Al requires 3 moles of MnO for reaction to proceed forward
how many moles of each are present?
calculate using
moles=given mass/molar mass

then see which is limiting if it were MnO then it wud have got completely used up remaing some aluminium
if u dont understand this find out the moles of Al and MnO
and type it here i will help u further

Answer 2

yes I did it in the above way...and i got the answer is 4.32 moles

Answer 3

nice

Answer 4

is it correct???
4.32 Al moles are left

Answer 5

i got 5.08 moles of Al

Answer 6

listen the limiting reagent is MnO so 5.7 moles will be used untill it lasts
now we se that
for 1 Al2O3 ---> 3 MnO
so for 1 MnO ---> 1/3 Al2O3

so for 5.6 MnO ---> 1.9 Al2O3

Answer 7

u shud never take the products ratio
wat u want is whether there is any reactant left out unreacted
if the reaction itself doesnt proceed in which courage did u go to check up with the pdts

first let the reaction take place and see how much of alumuminium is left behind that is uwat the problem requires

Answer 8

how did u get 5.08 Al

Answer 9

do it by same approach
2Al + 3 MnO
2 moles of Al requires 3 Mno
so calculate the moles

remeber while calculating moles=given mass/molar mass
molar mass is not same as atomic number