1.

Consider the following balanced chemical equation: 2A+4B_(2)+6C_(3)rarrP+2Q If initially 6.023xx 10^(24) atoms orf A, 448L of B_(2) gas at NTP and 960 gm of C_(3) gas are taken, which of the following is incorrect? [Given : Atomic mass of C=8]

Answer»

total number of species in final mixture is 4
total number of species in final mixture is 3
A and `B_(2)` both are present in limited amount amount.
`C_(3)` is in excess.

Solution :`n_(A)=10`
`n_(B)=20`
`n_(C)=(960)/(24)=40`
`2A+4B_(2)+6C_(3)rarrP+2Q`
Limiting reagent `rarrA & B_(2)`
so only 3 species will be left in final mixture.


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