Answer :
### Part A: Balancing the Chemical Equation
To balance the given chemical equation, let's start with the initial unbalanced equation:
[tex]\[ \_ P_4 + \ldots O_2 \rightarrow \_ P_2O_5 \][/tex]
1. Balancing Phosphorus (P):
- On the left side, consider we have 1 molecule of [tex]\( P_4 \)[/tex]. This means we have 4 phosphorus atoms.
- On the right side, 1 molecule of [tex]\( P_2O_5 \)[/tex] contains 2 phosphorus atoms.
To balance phosphorus, we need to have the same number of P atoms on both sides. Thus, we need 2 molecules of [tex]\( P_2O_5 \)[/tex] on the right to account for all 4 phosphorus atoms from one [tex]\( P_4 \)[/tex] molecule:
[tex]\[ P_4 + O_2 \rightarrow 2 P_2O_5 \][/tex]
2. Balancing Oxygen (O):
- On the left side, we have [tex]\( O_2 \)[/tex] molecules. Oxygen is diatomic in its element form, so each [tex]\( O_2 \)[/tex] molecule contains 2 oxygen atoms.
- On the right side, each [tex]\( P_2O_5 \)[/tex] molecule contains 5 oxygen atoms. Since we have 2 [tex]\( P_2O_5 \)[/tex] molecules, we have a total of 2 * 5 = 10 oxygen atoms.
To balance the oxygen atoms, we need 5 [tex]\( O_2 \)[/tex] molecules on the left side:
[tex]\[ P_4 + 5 O_2 \rightarrow 2 P_2O_5 \][/tex]
3. Balancing Phosphorus again:
Since we balanced phosphorus first, we need 2 molecules of [tex]\( P_4 \)[/tex] on the left to produce the necessary P atoms:
[tex]\[ 2 P_4 + 5 O_2 \rightarrow 4 P_2O_5 \][/tex]
Now the equation is balanced with 4 phosphorus atoms and 10 oxygen atoms on both sides.
### Final Balanced Equation:
[tex]\[ 2 P_4 + 5 O_2 \rightarrow 4 P_2O_5 \][/tex]
### Part B: Mass of [tex]\(P_2O_5\)[/tex] Produced from 496 grams of [tex]\(P_4\)[/tex]
We need to calculate how many grams of [tex]\( P_2O_5 \)[/tex] can be produced from 496 grams of [tex]\( P_4 \)[/tex].
1. Molar Mass Calculation:
- The molar mass of [tex]\( P_4 \)[/tex] is calculated by noting that each phosphorus atom (P) has an atomic mass of 31 grams/mole. Since there are 4 phosphorus atoms in [tex]\( P_4 \)[/tex]:
[tex]\[ \text{Molar mass of } P_4 = 4 \times 31 = 124 \text{ grams/mole} \][/tex]
- The molar mass of [tex]\( P_2O_5 \)[/tex] is calculated by noting there are 2 phosphorus atoms and 5 oxygen atoms. The atomic mass of oxygen (O) is 16 grams/mole:
[tex]\[ \text{Molar mass of } P_2O_5 = (2 \times 31) + (5 \times 16) = 62 + 80 = 142 \text{ grams/mole} \][/tex]
2. Calculating Moles of [tex]\( P_4 \)[/tex]:
- Given 496 grams of [tex]\( P_4 \)[/tex], the number of moles of [tex]\( P_4 \)[/tex] is:
[tex]\[ \text{Moles of } P_4 = \frac{496 \text{ grams}}{124 \text{ grams/mole}} = 4 \text{ moles} \][/tex]
3. Stoichiometry:
- From the balanced equation [tex]\( 2 P_4 + 5 O_2 \rightarrow 4 P_2O_5 \)[/tex], we see that 2 moles of [tex]\( P_4 \)[/tex] produce 4 moles of [tex]\( P_2O_5 \)[/tex]. Therefore, 1 mole of [tex]\( P_4 \)[/tex] produces 2 moles of [tex]\( P_2O_5 \)[/tex].
- Since we have 4 moles of [tex]\( P_4 \)[/tex]:
[tex]\[ \text{Moles of } P_2O_5 = 4 \text{ moles of } P_4 \][/tex]
4. Calculating Mass of [tex]\( P_2O_5 \)[/tex]:
- The mass of [tex]\( P_2O_5 \)[/tex] produced can be calculated by multiplying the moles of [tex]\( P_2O_5 \)[/tex] produced by its molar mass:
[tex]\[ \text{Mass of } P_2O_5 = 4 \text{ moles} \times 142 \text{ grams/mole} = 568 \text{ grams} \][/tex]
Therefore, 496 grams of [tex]\( P_4 \)[/tex] can produce 568 grams of [tex]\( P_2O_5 \)[/tex].
To balance the given chemical equation, let's start with the initial unbalanced equation:
[tex]\[ \_ P_4 + \ldots O_2 \rightarrow \_ P_2O_5 \][/tex]
1. Balancing Phosphorus (P):
- On the left side, consider we have 1 molecule of [tex]\( P_4 \)[/tex]. This means we have 4 phosphorus atoms.
- On the right side, 1 molecule of [tex]\( P_2O_5 \)[/tex] contains 2 phosphorus atoms.
To balance phosphorus, we need to have the same number of P atoms on both sides. Thus, we need 2 molecules of [tex]\( P_2O_5 \)[/tex] on the right to account for all 4 phosphorus atoms from one [tex]\( P_4 \)[/tex] molecule:
[tex]\[ P_4 + O_2 \rightarrow 2 P_2O_5 \][/tex]
2. Balancing Oxygen (O):
- On the left side, we have [tex]\( O_2 \)[/tex] molecules. Oxygen is diatomic in its element form, so each [tex]\( O_2 \)[/tex] molecule contains 2 oxygen atoms.
- On the right side, each [tex]\( P_2O_5 \)[/tex] molecule contains 5 oxygen atoms. Since we have 2 [tex]\( P_2O_5 \)[/tex] molecules, we have a total of 2 * 5 = 10 oxygen atoms.
To balance the oxygen atoms, we need 5 [tex]\( O_2 \)[/tex] molecules on the left side:
[tex]\[ P_4 + 5 O_2 \rightarrow 2 P_2O_5 \][/tex]
3. Balancing Phosphorus again:
Since we balanced phosphorus first, we need 2 molecules of [tex]\( P_4 \)[/tex] on the left to produce the necessary P atoms:
[tex]\[ 2 P_4 + 5 O_2 \rightarrow 4 P_2O_5 \][/tex]
Now the equation is balanced with 4 phosphorus atoms and 10 oxygen atoms on both sides.
### Final Balanced Equation:
[tex]\[ 2 P_4 + 5 O_2 \rightarrow 4 P_2O_5 \][/tex]
### Part B: Mass of [tex]\(P_2O_5\)[/tex] Produced from 496 grams of [tex]\(P_4\)[/tex]
We need to calculate how many grams of [tex]\( P_2O_5 \)[/tex] can be produced from 496 grams of [tex]\( P_4 \)[/tex].
1. Molar Mass Calculation:
- The molar mass of [tex]\( P_4 \)[/tex] is calculated by noting that each phosphorus atom (P) has an atomic mass of 31 grams/mole. Since there are 4 phosphorus atoms in [tex]\( P_4 \)[/tex]:
[tex]\[ \text{Molar mass of } P_4 = 4 \times 31 = 124 \text{ grams/mole} \][/tex]
- The molar mass of [tex]\( P_2O_5 \)[/tex] is calculated by noting there are 2 phosphorus atoms and 5 oxygen atoms. The atomic mass of oxygen (O) is 16 grams/mole:
[tex]\[ \text{Molar mass of } P_2O_5 = (2 \times 31) + (5 \times 16) = 62 + 80 = 142 \text{ grams/mole} \][/tex]
2. Calculating Moles of [tex]\( P_4 \)[/tex]:
- Given 496 grams of [tex]\( P_4 \)[/tex], the number of moles of [tex]\( P_4 \)[/tex] is:
[tex]\[ \text{Moles of } P_4 = \frac{496 \text{ grams}}{124 \text{ grams/mole}} = 4 \text{ moles} \][/tex]
3. Stoichiometry:
- From the balanced equation [tex]\( 2 P_4 + 5 O_2 \rightarrow 4 P_2O_5 \)[/tex], we see that 2 moles of [tex]\( P_4 \)[/tex] produce 4 moles of [tex]\( P_2O_5 \)[/tex]. Therefore, 1 mole of [tex]\( P_4 \)[/tex] produces 2 moles of [tex]\( P_2O_5 \)[/tex].
- Since we have 4 moles of [tex]\( P_4 \)[/tex]:
[tex]\[ \text{Moles of } P_2O_5 = 4 \text{ moles of } P_4 \][/tex]
4. Calculating Mass of [tex]\( P_2O_5 \)[/tex]:
- The mass of [tex]\( P_2O_5 \)[/tex] produced can be calculated by multiplying the moles of [tex]\( P_2O_5 \)[/tex] produced by its molar mass:
[tex]\[ \text{Mass of } P_2O_5 = 4 \text{ moles} \times 142 \text{ grams/mole} = 568 \text{ grams} \][/tex]
Therefore, 496 grams of [tex]\( P_4 \)[/tex] can produce 568 grams of [tex]\( P_2O_5 \)[/tex].
