4. What is the oxidizing agem 2Al^3+(aq)+2Fe(s)arrow 2Al(s) A) Al3+ B) Fe C) Al D) Fe^2+ E) none 5. Determine the limiting reactant (LR) and the mass (ing) of nitrogen that can be formed from 50.0 g N_(2)O_(4) and 45.0gN_(2)H_(4) from the unbalanced reaction given below. Some possibly useful atomic and molar masses are as follows: N=14.01g/mol H=1.008g/mol,O=16.000g/mol,N_(2)O_(4)=92.02g/mol,N_(2)H_(4)=32.05g/mol N_(2)O_(4)(l)+N_(2)H_(4)(l)arrow N_(2)(g)+H_(2)O(g) A) LR=N_(2)H_(4),59.0gN_(2) formed B) LR=N_(2)O_(4),45.7gN_(2) formed C) LR=N_(2)H_(4),13.3gN_(2) formed D) LR=N_(2)O_(4),105gN_(2)formed E) No LR, 45.0 g N_(2) formed

4. The correct answer is D) $Fe^{2+}$. In the given reaction, $Al^{3+}$ is reduced to $Al$ and $Fe$ is oxidized to $Fe^{2+}$. Therefore, the oxidizing agent is $Fe^{2+}$.<br /><br />5. To determine the limiting reactant (LR) and the mass of nitrogen that can be formed, we need to compare the mole ratio of the reactants with the stoichiometric ratio in the balanced equation.<br /><br />First, let's calculate the moles of $N_{2}O_{4}$ and $N_{2}H_{4}$:<br /><br />Moles of $N_{2}O_{4}$ = $\frac{50.0 \text{ g}}{92.02 \text{ g/mol}} = 0.543 \text{ mol}$<br /><br />Moles of $N_{2}H_{4}$ = $\frac{45.0 \text{ g}}{32.05 \text{ g/mol}} = 1.406 \text{ mol}$<br /><br />Next, let's compare the mole ratio of $N_{2}O_{4}$ to $N_{2}H_{4}$ with the stoichiometric ratio in the balanced equation:<br /><br />Stoichiometric ratio: 1 mole of $N_{2}O_{4}$ reacts with 1 mole of $N_{2}H_{4$ to produce 1 mole of $N_{2}$.<br /><br />Mole ratio: 0.543 mol of $N_{2}O_{4}$ reacts with 1.406 mol of $N_{2}H_{4}$.<br /><br />Since the mole ratio of $N_{2}O_{4}$ to $N_{2}H_{4}$ is less than the stoichiometric ratio, $N_{2}O_{4}$ is the limiting reactant.<br /><br />Now, let's calculate the mass of nitrogen that can be formed:<br /><br />Moles of $N_{2}$ formed = moles of limiting reactant = 0.543 mol<br /><br />Mass of nitrogen formed = moles of $N_{2}$ formed * molar mass of $N_{2}$<br /><br />Mass of nitrogen formed = 0.543 mol * 28.02 g/mol = 15.24 g<br /><br />Therefore, the correct answer is B) $LR=N_{2}O_{4},15.24gN_{2}$ formed.