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??
.......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.
I don't get what i have to do at all, or what the 1.20-2x=1.02 gives me?
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 = ?
To solve this problem, we need to apply the concepts of stoichiometry and use the given information about the initial and final amounts of H2.
1. Start by writing the balanced chemical equation for the reaction:
2H2 + 2NO -> N2 + 2H2O
2. Determine the stoichiometric ratio between H2 and NO.
From the equation, we see that 2 moles of H2 react with 2 moles of NO. Therefore, the ratio is 1:1, meaning that for every 2 moles of H2 used, 2 moles of NO will be used.
3. Calculate the amount of NO reacted.
We are given that initially, 1.20 moles of H2 were present, and after 5 seconds, the amount reduces to 1.02 moles. Thus, 1.20 - 1.02 = 0.18 moles of H2 have been consumed.
Since the ratio between H2 and NO is 1:1, this means that 0.18 moles of NO have also been consumed.
4. Calculate the amount of NO remaining.
Subtract the amount of NO consumed from the initial amount:
Initial amount of NO - consumed amount of NO = Remaining amount of NO
0.95 moles - 0.18 moles = 0.77 moles of NO remaining.
5. Calculate the amount of N2 and H2O produced.
Using the stoichiometric ratio from the balanced equation, we see that for every 2 moles of NO consumed, 1 mole of N2 and 2 moles of H2O are produced.
Therefore, for the consumption of 0.18 moles of NO, we have:
0.09 moles of N2 produced and 0.18 moles of H2O produced.
In summary:
- The amount of NO remaining is 0.77 moles.
- The amount of N2 produced is 0.09 moles.
- The amount of H2O produced is 0.18 moles.