Question

short answer ii: write the correct answer in the blank.

1. what unit is equivalent to avogadros number of objects?
2. what is a set of symbols that tells how many of each type of atom are found in a molecule?
3. what is the sum of the atomic masses of all the atoms in a molecule?
4. n₂h₄ (hydrazine) is sometimes used as a rocket fuel because it reacts readily with oxygen - containing compounds to produce large amounts of hot gas; one such reaction is 2 n₂h₄(l) + n₂o₄(l) → 3 n₂(g) + 4 h₂o(g). if 3 mol n₂h₄ and 4 mol n₂o₄ are available at the start of this reaction, which substance will be the limiting reactant?

Explanation:

Step1: Analyze the reaction equation

The reaction is \(2N_2H_4(l)+N_2O_4(l)\to3N_2(g) + 4H_2O(g)\). The mole ratio of \(N_2H_4\) to \(N_2O_4\) is \(2:1\).

Step2: Calculate the amount of \(N_2O_4\) needed for \(3\ mol\ N_2H_4\)

If we have \(n(N_2H_4)=3\ mol\), according to the mole ratio \(\frac{n(N_2H_4)}{n(N_2O_4)}=\frac{2}{1}\), then \(n(N_2O_4)\) needed is \(n(N_2O_4)=\frac{3\ mol}{2}=1.5\ mol\).

Step3: Compare the available amount of \(N_2O_4\)

We have \(n(N_2O_4)_{available} = 4\ mol\). Since \(1.5\ mol<4\ mol\), \(N_2H_4\) is not in excess.

Step4: Calculate the amount of \(N_2H_4\) needed for \(4\ mol\ N_2O_4\)

Using the mole ratio \(\frac{n(N_2H_4)}{n(N_2O_4)}=\frac{2}{1}\), if \(n(N_2O_4) = 4\ mol\), then \(n(N_2H_4)\) needed is \(n(N_2H_4)=2\times4\ mol = 8\ mol\).

Step5: Compare the available amount of \(N_2H_4\)

We have \(n(N_2H_4)_{available}=3\ mol\). Since \(3\ mol<8\ mol\), \(N_2H_4\) is the limiting reactant.

Answer:

\(N_2H_4\)