Consider the reaction: 2H2+2NO -> N2+2H2O. Initially 1.20mol of H2 and 0.95mol of NO are placed in a 2.50L container(all reactants and products are gases) if after 5seconds the amount of H2 has been reduced to 1.02mol.
How much NO is left? And how much N2 and H2O has been prodeced?
......2H2 + 2NO -> N2 + 2H2O
I.....1.2..0.95....0......0
C.....-2x...-2x....x.....2x
5sec..1.02.0.95-2x..x....2x
So 1.20-2x = 1.02
Solve for x and evaluate the other values you want to calculate.
To determine how much NO is left and how much N2 and H2O has been produced in the reaction, we need to use the stoichiometry of the reaction and the given information. Here's how to approach it step by step:
Step 1: Determine the limiting reactant
To find the limiting reactant, we compare the initial moles of both H2 and NO to the stoichiometry of the reaction. The balanced equation tells us that 2 moles of H2 react with 2 moles of NO to produce 1 mole of N2 and 2 moles of H2O.
For H2:
Initial moles of H2 = 1.20 mol
Moles of H2 remaining = 1.02 mol
For NO:
Initial moles of NO = 0.95 mol
To find the limiting reactant, we compare the moles of H2 and NO to the stoichiometry of the reaction. Since both H2 and NO have a stoichiometric coefficient of 2, we can compare them directly. In this case, the limiting reactant is the one that produces the smallest amount of product. So, the limiting reactant is H2.
Step 2: Determine the moles of other reactants and products
Since H2 is the limiting reactant, we can use the stoichiometry to determine the moles of NO, N2, and H2O.
From the balanced equation: 2H2 + 2NO -> N2 + 2H2O
Since the stoichiometric coefficient for H2O is 2, we know that the number of moles of H2O produced is twice the number of moles of H2 consumed.
Moles of H2O produced = 2 × (initial moles of H2 - moles of H2 remaining)
= 2 × (1.20 - 1.02)
= 0.36 mol
Since the stoichiometric coefficient for N2 is 1, we know that the number of moles of N2 produced is the same as the number of moles of H2 consumed.
Moles of N2 produced = initial moles of H2 - moles of H2 remaining
= 1.20 - 1.02
= 0.18 mol
Step 3: Calculate the moles of NO remaining
From the stoichiometry of the reaction, the moles of NO consumed is the same as the moles of H2 consumed because the stoichiometric coefficients are the same.
Moles of NO consumed = initial moles of H2 - moles of H2 remaining
= 1.20 - 1.02
= 0.18 mol
Therefore, the moles of NO remaining is the initial moles of NO minus the moles of NO consumed.
Moles of NO remaining = initial moles of NO - moles of NO consumed
= 0.95 - 0.18
= 0.77 mol
So, the amount of NO remaining is 0.77 mol.
To summarize:
- The amount of NO remaining is 0.77 mol.
- The amount of N2 produced is 0.18 mol.
- The amount of H2O produced is 0.36 mol.