Answer :
Let's determine the limiting reactant and the mass of phosphoric acid, [tex]\( H_3PO_4 \)[/tex], produced in the reaction.
Step 1: Calculate moles of each reactant.
- For [tex]\( P_4O_{10} \)[/tex]:
The molar mass of [tex]\( P_4O_{10} \)[/tex] is 283.88 g/mol.
Mass given = 82.3 g.
Moles of [tex]\( P_4O_{10} \)[/tex] = [tex]\(\frac{82.3 \text{ g}}{283.88 \text{ g/mol}} \approx 0.290 \text{ moles}\)[/tex].
- For [tex]\( H_2O \)[/tex]:
The molar mass of [tex]\( H_2O \)[/tex] is 18.02 g/mol.
Mass given = 66.9 g.
Moles of [tex]\( H_2O \)[/tex] = [tex]\(\frac{66.9 \text{ g}}{18.02 \text{ g/mol}} \approx 3.713 \text{ moles}\)[/tex].
Step 2: Use the balanced equation to find the stoichiometric relationship.
The balanced equation is:
[tex]\[ P_4O_{10} + 6 H_2O \longrightarrow 4 H_3PO_4 \][/tex]
From this, 1 mole of [tex]\( P_4O_{10} \)[/tex] reacts with 6 moles of [tex]\( H_2O \)[/tex].
Step 3: Calculate the moles of [tex]\( H_2O \)[/tex] needed for the reaction.
- Moles of [tex]\( H_2O \)[/tex] needed = moles of [tex]\( P_4O_{10} \times 6 \)[/tex]
- Moles of [tex]\( H_2O \)[/tex] needed = [tex]\( 0.290 \text{ moles} \times 6 = 1.739 \text{ moles} \)[/tex]
Step 4: Determine the limiting reactant.
Compare the moles of [tex]\( H_2O \)[/tex] available with the moles needed:
- Moles of [tex]\( H_2O \)[/tex] available = 3.713 moles.
- Moles of [tex]\( H_2O \)[/tex] needed = 1.739 moles.
Since we have more [tex]\( H_2O \)[/tex] available than needed, [tex]\( P_4O_{10} \)[/tex] is the limiting reactant.
Step 5: Calculate the moles of [tex]\( H_3PO_4 \)[/tex] produced.
From the balanced equation, 1 mole of [tex]\( P_4O_{10} \)[/tex] produces 4 moles of [tex]\( H_3PO_4 \)[/tex].
- Moles of [tex]\( H_3PO_4 \)[/tex] produced = moles of [tex]\( P_4O_{10} \times 4 \)[/tex]
- Moles of [tex]\( H_3PO_4 \)[/tex] produced = [tex]\( 0.290 \text{ moles} \times 4 = 1.160 \text{ moles} \)[/tex]
Step 6: Calculate the mass of [tex]\( H_3PO_4 \)[/tex] produced.
The molar mass of [tex]\( H_3PO_4 \)[/tex] is 98.00 g/mol.
- Mass of [tex]\( H_3PO_4 \)[/tex] = moles of [tex]\( H_3PO_4 \times 98.00 \text{ g/mol} \)[/tex]
- Mass of [tex]\( H_3PO_4 \)[/tex] = [tex]\( 1.160 \text{ moles} \times 98.00 \text{ g/mol} \approx 113.65 \text{ g} \)[/tex]
So, the limiting reactant is [tex]\( P_4O_{10} \)[/tex] and the mass of [tex]\( H_3PO_4 \)[/tex] produced is approximately 113.65 g.
Step 1: Calculate moles of each reactant.
- For [tex]\( P_4O_{10} \)[/tex]:
The molar mass of [tex]\( P_4O_{10} \)[/tex] is 283.88 g/mol.
Mass given = 82.3 g.
Moles of [tex]\( P_4O_{10} \)[/tex] = [tex]\(\frac{82.3 \text{ g}}{283.88 \text{ g/mol}} \approx 0.290 \text{ moles}\)[/tex].
- For [tex]\( H_2O \)[/tex]:
The molar mass of [tex]\( H_2O \)[/tex] is 18.02 g/mol.
Mass given = 66.9 g.
Moles of [tex]\( H_2O \)[/tex] = [tex]\(\frac{66.9 \text{ g}}{18.02 \text{ g/mol}} \approx 3.713 \text{ moles}\)[/tex].
Step 2: Use the balanced equation to find the stoichiometric relationship.
The balanced equation is:
[tex]\[ P_4O_{10} + 6 H_2O \longrightarrow 4 H_3PO_4 \][/tex]
From this, 1 mole of [tex]\( P_4O_{10} \)[/tex] reacts with 6 moles of [tex]\( H_2O \)[/tex].
Step 3: Calculate the moles of [tex]\( H_2O \)[/tex] needed for the reaction.
- Moles of [tex]\( H_2O \)[/tex] needed = moles of [tex]\( P_4O_{10} \times 6 \)[/tex]
- Moles of [tex]\( H_2O \)[/tex] needed = [tex]\( 0.290 \text{ moles} \times 6 = 1.739 \text{ moles} \)[/tex]
Step 4: Determine the limiting reactant.
Compare the moles of [tex]\( H_2O \)[/tex] available with the moles needed:
- Moles of [tex]\( H_2O \)[/tex] available = 3.713 moles.
- Moles of [tex]\( H_2O \)[/tex] needed = 1.739 moles.
Since we have more [tex]\( H_2O \)[/tex] available than needed, [tex]\( P_4O_{10} \)[/tex] is the limiting reactant.
Step 5: Calculate the moles of [tex]\( H_3PO_4 \)[/tex] produced.
From the balanced equation, 1 mole of [tex]\( P_4O_{10} \)[/tex] produces 4 moles of [tex]\( H_3PO_4 \)[/tex].
- Moles of [tex]\( H_3PO_4 \)[/tex] produced = moles of [tex]\( P_4O_{10} \times 4 \)[/tex]
- Moles of [tex]\( H_3PO_4 \)[/tex] produced = [tex]\( 0.290 \text{ moles} \times 4 = 1.160 \text{ moles} \)[/tex]
Step 6: Calculate the mass of [tex]\( H_3PO_4 \)[/tex] produced.
The molar mass of [tex]\( H_3PO_4 \)[/tex] is 98.00 g/mol.
- Mass of [tex]\( H_3PO_4 \)[/tex] = moles of [tex]\( H_3PO_4 \times 98.00 \text{ g/mol} \)[/tex]
- Mass of [tex]\( H_3PO_4 \)[/tex] = [tex]\( 1.160 \text{ moles} \times 98.00 \text{ g/mol} \approx 113.65 \text{ g} \)[/tex]
So, the limiting reactant is [tex]\( P_4O_{10} \)[/tex] and the mass of [tex]\( H_3PO_4 \)[/tex] produced is approximately 113.65 g.
