N2+2H2=N2H4 What is the concentration of H2 at equalibrium when (N2)=.033 and (N2H2)= .042 M? Keq=5.29

Answer 1

..........N2 + 2H2 ==> N2H4
aquil..0.033....x......0.042

Keq = (N2H2)/(N2)(H2)^2
Substitute and solve for x.

Answer 2

.04905 M

Answer 3

0.4905 M

Answer 4

Yes but if your prof is picky about the number of significant figures s/he may not give you full credit for that answer.

Answer 5

ok thanks!