Question

1. how much energy is required to decompose 755 mg of pcl₃, according to the reaction below? pcl₃ = 137.32 g/mol 4pcl₃(g) → p₄(s) + 6cl₂(g)δhᵣ° = +1207 kj 549.28 = 1207 0.6028 3) how much energy is evolved during the formation of 35.8 g of fe, according to the reaction below? fe₂o₃(s) + 2al(s) → al₂o₃(s) + 2fe(s) δhᵣ° = -852 kj

Explanation:

Step1: Convert mass of $PCl_3$ to moles

First, convert 755 mg of $PCl_3$ to grams ($755\ mg=0.755\ g$). Then use the molar - mass formula $n = \frac{m}{M}$, where $m = 0.755\ g$ and $M = 137.32\ g/mol$. So, $n=\frac{0.755\ g}{137.32\ g/mol}\approx0.0055\ mol$.

Step2: Determine energy based on stoichiometry

From the reaction $4PCl_3(g)\to P_4(s)+6Cl_2(g)\ \Delta H_r^{\circ}= + 1207\ kJ$, 4 moles of $PCl_3$ require 1207 kJ of energy. Let $x$ be the energy required for 0.0055 mol of $PCl_3$. Set up a proportion: $\frac{x}{0.0055\ mol}=\frac{1207\ kJ}{4\ mol}$. Solving for $x$, we get $x=\frac{1207\ kJ\times0.0055\ mol}{4\ mol}\approx1.66\ kJ$.

for second part:

Step1: Convert mass of Fe to moles

Use the molar - mass of Fe, which is $M_{Fe}=55.85\ g/mol$. Given $m = 35.8\ g$, then $n=\frac{m}{M}=\frac{35.8\ g}{55.85\ g/mol}\approx0.641\ mol$.

Step2: Determine energy based on stoichiometry

From the reaction $Fe_2O_3(s)+2Al(s)\to Al_2O_3(s)+2Fe(s)\ \Delta H_r^{\circ}=-852\ kJ$, 2 moles of Fe formation evolves 852 kJ of energy. Let $y$ be the energy evolved for 0.641 mol of Fe. Set up a proportion: $\frac{y}{0.641\ mol}=\frac{852\ kJ}{2\ mol}$. Solving for $y$, we get $y=\frac{852\ kJ\times0.641\ mol}{2\ mol}\approx273\ kJ$.

Answer:

1.66 kJ