1.04 g H2 reacts with 10.2 g N2 producing 2.71 g NH3, what is the theoretical yield in grams?
To determine the theoretical yield of NH3 in this reaction, we need to identify the limiting reactant. The limiting reactant is the one that is completely consumed, thereby limiting the amount of product that can be formed.
Let's start by calculating the moles of each reactant.
1 mole of H2 weighs 2 grams, so 1.04 g of H2 is equal to (1.04 g / 2 g/mol) = 0.52 moles of H2.
1 mole of N2 weighs 28 grams, so 10.2 g of N2 is equal to (10.2 g / 28 g/mol) ≈ 0.364 moles of N2.
Next, we need to determine the stoichiometry of the reaction to find the mole ratio of H2 to NH3 and N2 to NH3. From the balanced equation:
N2 + 3H2 -> 2NH3
We can see that for each mole of N2, we need 3 moles of H2 to produce 2 moles of NH3.
Now, let's calculate the moles of NH3 that can be formed from each reactant:
From H2: 0.52 moles of H2 * (2 moles of NH3 / 3 moles of H2) = 0.347 moles of NH3
From N2: 0.364 moles of N2 * (2 moles of NH3 / 1 mole of N2) = 0.728 moles of NH3
We can see that the limiting reactant is H2 because it produces the least amount of NH3. Therefore, the maximum amount of NH3 that can be produced is 0.347 moles.
Finally, let's convert the moles of NH3 to grams:
0.347 moles of NH3 * (17 g/mol) = 5.90 g
Therefore, the theoretical yield of NH3 in grams is 5.90 g.