Calculate how many moles of NH3 form when 3.55 moles of N2H4 completely react according to the equation: 3N2H4 (l) ---> 4NH3 (g) + N2 (g) ?

What we're doing here is calculating basic mole-mole relationships, something that you'll be doing quite a bit!

The steps to solving mole-mole problems like this are

write the balanced chemical equation for the reaction (this is given)

Using simple dimensional analysis, it looks like this:

To calculate the number of moles of NH₃ formed when 3.55 moles of N₂H₄ completely react, we first use the stoichiometric coefficients from the balanced equation.

Given: 3N₂H₄ (l) → 4NH₃ (g) + N₂ (g)

Since the stoichiometric coefficient of N₂H₄ is 3, and the coefficient of NH₃ is 4, we can set up a ratio:

(4 moles NH₃ / 3 moles N₂H₄) * 3.55 moles N₂H₄ = 4.73 moles NH₃

So, 4.73 moles of NH₃ would form when 3.55 moles of N₂H₄ completely react.

To solve this problem, we'll use the stoichiometry of the reaction.

Given: Number of moles of N2H4 (l) = 3.55 moles

From the balanced chemical equation: 3N2H4 (l) ---> 4NH3 (g) + N2 (g)

We can see that 3 moles of N2H4 produce 4 moles of NH3.

Using the ratio of moles of N2H4 to moles of NH3: 1 mole of N2H4 produces ( \frac{4}{3} ) moles of NH3.

Therefore, for 3.55 moles of N2H4: [ 3.55 \text{ moles N2H4} \times \frac{4}{3} \text{ moles NH3/mol N2H4} = 4.733 \text{ moles NH3} ]

So, 3.55 moles of N2H4 produce 4.733 moles of NH3.