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@Dexter
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.@Gogeta Hlo
You see You need 1 Mole of N2 for it to react with 3 moles of H2.
1 mole at stp is 22.4L
Therefore 1 mole of N2, that is 22.4L of N2 reacts with 67.2 Litres of H2.
However since he has given only 2L of N2
Then 22.4 L of N2 needs 67.2L of H2
2L of N2 ---------------- ?
2*67.2/22.4 which gives 6 Litres of H2.
however H2 is only 2L here and is present at a small amt, so it will get used up quickly leaving some Nitrogen behind thus becoming the limiting reagent.
tyy got it
alr np gotchu man.
1 mole of nitrogen needs 3 moles of hydrogen to react completely.
So x moles of nitrogen will need 3x moles of hydrogen to react completely
In the same way 2 moles of nitrogen would need 6 moles (3*2) of hydrogen to react completely
.
As only 2 moles of hydrogen are present so it's the limiting reagent,
you can also divide the available moles by the stoichiometric coefficients and compare the values, the smaller value is the limiting reagent
+solved @Flint
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