To find the number of moles of N2 that reacted when 0.65 moles of NH3 is produced, we can use the stoichiometry of the reaction.
From the balanced chemical equation, we can see that 1 mole of N2 reacts to produce 2 moles of NH3.
So, if x moles of N2 reacts, it will produce (2x) moles of NH3.
Now, we can set up a proportion to solve for x:
(2x moles NH3 / 1 mole N2) = (0.65 moles NH3 / x mole N2)
Cross-multiplying the proportion, we get:
2x = 0.65
Dividing both sides by 2, we find:
x = 0.65 / 2
x = 0.325 moles of N2
Therefore, 0.325 moles of N2 reacted to produce 0.65 moles of NH3.
To find the number of moles of NH3 produced when 1.2 moles of H2 reacts, we can again use the stoichiometry of the reaction.
From the balanced chemical equation, we can see that 3 moles of H2 react to produce 2 moles of NH3.
So, if y moles of H2 react, it will produce (2/3 * y) moles of NH3.
Now, we can set up a proportion to solve for y:
((2/3) * y moles NH3 / 3 moles H2) = (x moles NH3 / 1.2 moles H2)
Cross-multiplying the proportion, we get:
(2/3) * y = (1.2 * x)
Dividing both sides by (2/3), we find:
y = (1.2 * x) / (2/3)
Simplifying further, we get:
y = (1.2 * x) * (3/2)
y = (1.8 * x)
Therefore, 1.8 times the number of moles of H2 is the number of moles of NH3 produced.
So, when 1.2 moles of H2 reacts, it will produce:
(1.8 * 1.2) moles of NH3 = 2.16 moles of NH3