The reaction of NH3 and O2 forms NO and water. The NO can be used to convert P4 to P4O6 , forming N2 in the process. The P4O6 can be treated with water to form H3PO3 , which forms PH3 and H3PO4 when heated. Find the mass of PH3 that forms from the reaction of 1.30 g NH3 .

To solve the problem, we need to follow the series of reactions and convert the amount of ammonia (NH₃) given (1.30 g) into the amount of phosphine (PH₃) formed, step by step.

### Step 1: Write the balanced chemical equations for each reaction.

#### Reaction 1: Combustion of NH₃
\[ 4 NH_3 + 5 O_2 \rightarrow 4 NO + 6 H_2O \]
Here, for every 4 moles of NH₃, 4 moles of NO are formed.

#### Reaction 2: Reaction of NO with P₄ to form P₄O₆
\[ 6 NO + P_4 \rightarrow 2 N_2 + P_4O_6 \]
For every 6 moles of NO, 2 moles of N₂ and 1 mole of P₄O₆ are formed.

#### Reaction 3: Hydrolysis of P₄O₆ to form H₃PO₃
\[ P_4O_6 + 6 H_2O \rightarrow 4 H_3PO_3 \]
For every 1 mole of P₄O₆, 4 moles of H₃PO₃ are formed.

#### Reaction 4: Decomposition of H₃PO₃ into PH₃ and H₃PO₄ when heated
\[ 4 H_3PO_3 \rightarrow PH_3 + 3 H_3PO_4 \]
For every 4 moles of H₃PO₃, 1 mole of PH₃ is produced.

### Step 2: Convert the mass of NH₃ to moles
First, calculate the molar mass of NH₃:
\[
\text{Molar mass of NH₃} = 14.01 \, \text{(N)} + 3 \times 1.008 \, \text{(H)} = 17.034 \, \text{g/mol}
\]
Now, find the number of moles of NH₃:
\[
\text{Moles of NH₃} = \frac{1.30 \, \text{g}}{17.034 \, \text{g/mol}} = 0.0763 \, \text{mol}
\]

### Step 3: Use stoichiometry to calculate moles of PH₃ formed

1. From Reaction 1, for every 4 moles of NH₃, 4 moles of NO are formed. So, 0.0763 moles of NH₃ will produce the same amount of NO (since it's a 1:1 ratio):
   \[
   \text{Moles of NO} = 0.0763 \, \text{mol}
   \]

2. From Reaction 2, 6 moles of NO produce 1 mole of P₄O₆. Therefore:
   \[
   \text{Moles of P₄O₆} = \frac{0.0763 \, \text{mol NO}}{6} = 0.01272 \, \text{mol}
   \]

3. From Reaction 3, 1 mole of P₄O₆ produces 4 moles of H₃PO₃, so:
   \[
   \text{Moles of H₃PO₃} = 0.01272 \, \text{mol P₄O₆} \times 4 = 0.05088 \, \text{mol}
   \]

4. From Reaction 4, 4 moles of H₃PO₃ produce 1 mole of PH₃. So:
   \[
   \text{Moles of PH₃} = \frac{0.05088 \, \text{mol H₃PO₃}}{4} = 0.01272 \, \text{mol}
   \]

### Step 4: Convert moles of PH₃ to mass
Finally, calculate the molar mass of PH₃:
\[
\text{Molar mass of PH₃} = 30.97 \, \text{(P)} + 3 \times 1.008 \, \text{(H)} = 34.00 \, \text{g/mol}
\]
Now, calculate the mass of PH₃:
\[
\text{Mass of PH₃} = 0.01272 \, \text{mol} \times 34.00 \, \text{g/mol} = 0.4325 \, \text{g}
\]

### Final Answer:
The mass of PH₃ formed from the reaction of 1.30 g of NH₃ is **0.433 g** (rounded to three significant figures).

---

Would you like more details or further clarification?

Here are 5 related questions to consider:
1. What is the role of NO in converting P₄ to P₄O₆?
2. How does the stoichiometry of NH₃ to NO relate to real-world industrial applications?
3. Can the reaction be modified to increase PH₃ yield?
4. What would happen if P₄O₁₀ were formed instead of P₄O₆?
5. How does PH₃ compare to NH₃ in terms of molecular geometry?

**Tip:** Always make sure your chemical equations are balanced before performing stoichiometric calculations!

The reaction of NH3 and O2 forms NO and water. The NO can be used to convert P4 to P4O6, forming N2 in the process. The P4O6 can be treated with water to form H3PO3, which forms PH3 and H3PO4 when heated. Find the mass of PH3 that forms from the reaction of 1.30 g NH3.

Discover how to find the mass of PH3 formed from the reaction of 1.30 g of NH3 using stoichiometry. This detailed explanation covers balanced chemical equations, molar mass calculations, and step-by-step solution methods, ideal for high school chemistry students.

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