Question

Question was provided via image upload.

Explanation:

Step1: Balance reaction 1

Balance $N_2 + O_2
ightarrow NO_2$:
$1N_2 + 2O_2
ightarrow 2NO_2$
Mole ratio: $N_2:O_2:NO_2 = 1:2:2$

Step2: Balance reaction 2

Balance $SO_2 + O_2
ightarrow SO_3$:
$2SO_2 + 1O_2
ightarrow 2SO_3$
Mole ratio: $SO_2:O_2:SO_3 = 2:1:2$

Step3: Solve reaction 3

Use mole ratio $P_4:Cl_2 = 1:6$
$\text{Moles of } Cl_2 = 3 \times 6 = 18$

Step4a: Find limiting reactant for Q4

Reaction: $P_4 + 6Cl_2
ightarrow 4PCl_3$
Mole ratio $P_4:Cl_2 = 1:6$
$\text{Required } Cl_2 \text{ for 5 mol } P_4 = 5 \times 6 = 30$ mol. Only 22 mol $Cl_2$ available, so $Cl_2$ is limiting.
$\text{Moles of } PCl_3 = 22 \times \frac{4}{6} = \frac{44}{3} \approx 14.67$

Step4b: Find excess $P_4$

$\text{Moles of } P_4 \text{ reacted} = 22 \times \frac{1}{6} = \frac{11}{3}$
$\text{Excess } P_4 = 5 - \frac{11}{3} = \frac{4}{3} \approx 1.33$

Step4c: Check excess $Cl_2$

$Cl_2$ is limiting, so 0 moles left.

Step5: Identify limiting reactant (Q5)

Reaction: $2Al + 3Br_2
ightarrow 2AlBr_3$
Mole ratio $Al:Br_2 = 2:3$
$\text{Required } Br_2 \text{ for 3 mol } Al = 3 \times \frac{3}{2} = 4.5$ mol. Only 4 mol $Br_2$ available, so $Br_2$ is limiting.

Step5a: Identify excess reactant

$Al$ is the excess reactant.

Step5b: Calculate $AlBr_3$ produced

Use limiting $Br_2$:
$\text{Moles of } AlBr_3 = 4 \times \frac{2}{3} = \frac{8}{3} \approx 2.67$

Step5c: Find excess $Al$

$\text{Moles of } Al \text{ reacted} = 4 \times \frac{2}{3} = \frac{8}{3}$
$\text{Excess } Al = 3 - \frac{8}{3} = \frac{1}{3} \approx 0.33$

Answer:

1. Mole ratio: $N_2:O_2:NO_2 = 1:2:2$
2. Mole ratio: $SO_2:O_2:SO_3 = 2:1:2$
3. 18 moles
4. a. $\frac{44}{3}$ (≈14.67) moles

b. $\frac{4}{3}$ (≈1.33) moles
c. 0 moles

1. Limiting reactant: $Br_2$

a. Excess reactant: $Al$
b. $\frac{8}{3}$ (≈2.67) moles
c. $\frac{1}{3}$ (≈0.33) moles