To determine which reactant is in excess and how much tetraphosphorus trisulfide (P₄S₃) can be formed from the reaction of phosphorus and sulfur, we first need to analyze the balanced chemical reaction for the formation of P₄S₃:
The balanced equation for the formation of P₄S₃ from phosphorus (P) and sulfur (S) is:
\[ 4P + 6S \rightarrow P_4S_3 \]
From the equation, we can see the molar ratio of phosphorus to sulfur required to produce P₄S₃ is:
- 4 moles of P for every 6 moles of S.
Step 1: Calculate moles of each reactant
Molar masses:
- Molar mass of phosphorus (P) = 30.97 g/mol
- Molar mass of sulfur (S) = 32.07 g/mol
For Phosphorus:
\[ \text{Moles of } P = \frac{\text{Mass of } P}{\text{Molar mass of } P} = \frac{12.0 \text{ g}}{30.97 \text{ g/mol}} \approx 0.387 \text{ moles of } P \]
For Sulfur:
\[ \text{Moles of } S = \frac{\text{Mass of } S}{\text{Molar mass of } S} = \frac{10.0 \text{ g}}{32.07 \text{ g/mol}} \approx 0.311 \text{ moles of } S \]
Step 2: Determine the limiting reactant
Using the stoichiometry of the reaction:
- From the balanced equation, 4 moles of P reacts with 6 moles of S, which simplifies to a ratio of \( \frac{2}{3} \) or \( \frac{P}{S} = \frac{2}{3} \).
To find the limiting reactant, we can compare the actual ratios of available moles:
Using the ratio of available moles:
- For 0.387 moles of P, the required moles of S are:
\[ \text{Required moles of } S = 0.387 \text{ moles } P \times \frac{6 \text{ moles } S}{4 \text{ moles } P} = 0.581 \text{ moles } S \]
Since we have only 0.311 moles of S, sulfur (S) is the limiting reactant.
Step 3: Calculate the excess reactant
Now, we need to calculate how much phosphorus is left over after the reaction.
From the balanced reaction, for every 6 moles of S, you need 4 moles of P. Therefore, for 0.311 moles of S, the amount of P required is:
\[ \text{Required moles of } P = 0.311 \text{ moles } S \times \frac{4 \text{ moles } P}{6 \text{ moles } S} = 0.207 \text{ moles } P \]
Now we can calculate the mass of P consumed:
\[ \text{Mass of } P = 0.207 \text{ moles } P \times 30.97 \text{ g/mol} \approx 6.42 \text{ g of } P \]
Now subtract this from the initial mass of phosphorus to find the excess:
\[ \text{Excess P} = 12.0 \text{ g} - 6.42 \text{ g} = 5.58 \text{ g} \]
Step 4: Calculate the mass of P₄S₃ formed
From the stoichiometry of the balanced equation, we see that:
- 4 moles of P₄S₃ are produced for every 4 moles of P reacted.
The number of moles of P₄S₃ formed will be equal to the number of moles of S consumed (because S is the limiting reactant):
\[ \text{Moles of } P_4S_3 \text{ formed} = \frac{1}{6} \text{ moles of S consumed} = \frac{1}{6} \times 0.311 = 0.052 \text{ moles of } P_4S_3 \]
Now, calculate the mass of P₄S₃ formed:
The molar mass of P₄S₃: \[ \text{Molar mass of } P_4S_3 = (4 \times 30.97) + (3 \times 32.07) = 123.88 + 96.21 = 220.09 \text{ g/mol} \]
\[ \text{Mass of } P_4S_3 = \text{moles} \times \text{molar mass} = 0.052 \text{ moles} \times 220.09 \text{ g/mol} \approx 11.44 \text{ g} \]
Summary:
- The limiting reactant is Sulfur (S) in excess of Phosphorus (P) by approximately 5.58 grams.
- The mass of P₄S₃ formed is approximately 11.44 grams.
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