Respuesta :
Answer:
a) O2 is the limiting reactant
b) 5.75 grams P4O10
c) 5.79 grams P4O6
Explanation:
Step 1: Data given
Mass of P4 = 5.77 grams
Mass of O2 = 5.77 grams
Molar mass of P4 = 123.90 g/mol
Molar mass O2 = 32.0 g/mol
Step 2: The balanced equation
P4 + 3O2 → P4O6
Step 3: Calculate moles of P4
Moles P4 = mass P4 / molar mass P4
Moles P4 = 5.77 grams / 123.90 g/mol
Moles P4 = 0.0466 moles
Step 4: Calculate moles O2
Moles O2 = mass O2 / molar mass O2
Moles O2 = 5.77 grams / 32.0 g/mol
Moles O2 = 0.1803 moles
Step 5: Calculate limiting reactant
P4 is the limiting reactant in this reaction. It will completely be consumed (0.0466 moles). O2 is in excess, there will react 3*0.0466 = 0.1398 moles
There will remain 0.1803 - 0.1398 = 0.0405 moles O2
Step 6: Calculate the amount of P4O6
For 1 mol P4 we'll have 1 mol P4O6
For 0.0466 moles P4 we'll have 0.0466 moles P4O6
Step 7: The balanced equatio
P4O6 + 2O2 → P4O10
We have 0.0466 moles P4O6 and 0.0405 moles O2
Step 8: Calculate the limiting reactant
For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10
O2 is the limiting reactant. It will completely be consumed (0.0405 moles)
P4O6 is in excess. There will react 0.0405/2 = 0.02025 moles
There will remain 0.0466 - 0.02025 = 0.02635 moles P4O6
This is 0.02635 * 219.88 g/mol = 5.79 grams P4O6
Step 9: Calculate moles and mass of P4O10
For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10
For 0.0405 moles O2 we'll have 0.02025 moles P4O10
This is 0.02025 * 283.89 g/mol = 5.75 grams P4O10
a. The limiting reagent for the formation of P₄O₁₀ is O₂
b. The mass of P₄O₁₀ produced is 5.79 g
c. The mass of excess reactant left is 5.76138 g
Firstly, the limiting reagent is the reactant that is completely used up in a reaction while the excess reagent is the reactant that is not used up when the reaction is concluded.
Let's represents the series of the reactions chemically and balance them.
P₄ + O₂→ P₄O₆
P₄ + 3O₂→ P₄O₆
P₄O₆ + O₂→P₄O₁₀
P₄O₆ +2O₂→P₄O₁₀
Let's calculate the values to get the moles/mass of P₄O₆ that was used in the final chemical equation.
Therefore,
Moles of P₄ = mass/molar mass =5.77/123.90 = 0.04656981436≈0.0465 moles
Moles of O₂ = 5.77/32 = 0.1803125≈0.1803 moles
From the first chemical reaction, P₄ is the limiting reagent while O₂ is the excess reactant.
This means 0.0465 moles of P₄ reacted with 3 × 0.0465 moles of O₂(0.1395
moles)
The excess oxygen remaining in the first reaction will be 0.1803 - 0.1395 = 0.0408 moles
P₄O₆ +2O₂→P₄O₁₀
1 mole of P₄O₆ + 2 moles of O₂ = 1 mole of P₄O₁₀
0.0408 moles of O₂ will produce 0.0204 moles of P₄O₁₀
a.
The limiting reagent in the formation of P₄O₁₀ is O₂(0.0408 moles).
b.
The mass of of P₄O₁₀ produced can be below
moles = mass/molar mass =
mass = moles × molar mass
mass = 0.0204 × 283.90 = 5.79 grams
c.
The excess reactant(P₄O₆) mass will be
1 mole of P₄O₆ needs 2 moles of O₂ to produce 1 mole of P₄O₁₀
Therefore,
moles of P₄O₆ that will react will be = 0.0408/2 = 0.0204 moles
The remainder will be = 0.0466 - 0.0204 = 0.0262 moles
mass of the excess reactant left = 0.0262 × 219.9(moles × molar mass of P₄O₆) = 5.76138 g left.
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