Consider the reaction: 2H2+2NO-N2+2H2O. Initially 1.20mol of H2 and 0.95 mol of NO are placed in a container(all reactants and products are gases). If after 5 seconds the amount of H2 has been reduced to 1.02mol then how much NO is left and how much N2 and H2O has been produced??

Question

Consider the reaction: 2H2+2NO->N2+2H2O. Initially 1.20mol of H2 and 0.95 mol of NO are placed in a container(all reactants and products are gases). If after 5 seconds the amount of H2 has been reduced to 1.02mol then how much NO is left and how much N2 and H2O has been produced??

DrBob222 · Answer

.......2H2 + 2NO->N2 + 2H2O.
I.....1.20..0.95..0.....0
+5sec.-2x....-2x..x....2x
final.1.20-2x.0.95-2x..x...2x

The problem tells you that 1.20-2x = 1.02. Solve for x and evaluate the other values. Post your work if you get stuck.

Sean · Answer

I don't get what i have to do at all, or what the 1.20-2x=1.02 gives me?

DrBob222 · Answer

You don't? 1.20-2x = 1.02 -2x = 1.02-1.20 = -0.18 2x = 0.18 x = 0.18/2 = 0.09 So N2 = x = 0.09 H2O is 2x = 2*0.09 = 0.18 N2 left is 0.95-2x = ?