Sure, let's go through the problem step-by-step to calculate the mass of sodium that will react with 230 g of oxygen to form sodium oxide.
First, we need to consider the given reaction:
[tex]\[ 4Na + O_2 \rightarrow 2Na_2O \][/tex]
1. Molar Masses:
- Molar mass of [tex]\( Na \)[/tex] (Sodium) is [tex]\( 22.99 \)[/tex] g/mol.
- Molar mass of [tex]\( O_2 \)[/tex] (Oxygen gas) is [tex]\( 32.00 \)[/tex] g/mol.
- Molar mass of [tex]\( Na_2O \)[/tex] (Sodium Oxide) is calculated as:
[tex]\[ 2(22.99) + 16.00 = 45.98 + 16.00 = 61.98 \) g/mol.
2. Calculate Moles of \( O_2 \):
- Given the mass of \( O_2 \) as \( 230 \) g, we can calculate the moles of \( O_2 \) using the formula:
\[ \text{Moles of } O_2 = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} = \frac{230 \, \text{g}}{32.00 \, \text{g/mol}} \approx 7.1875 \, \text{moles} \][/tex]
3. Calculate Moles of [tex]\( Na \)[/tex] Needed:
- From the balanced equation, [tex]\( 4 \)[/tex] moles of [tex]\( Na \)[/tex] react with [tex]\( 1 \)[/tex] mole of [tex]\( O_2 \)[/tex]. So, the moles of sodium needed can be calculated as:
[tex]\[ \text{Moles of } Na = 7.1875 \, \text{moles of } O_2 \times 4 \, \text{moles of } Na/\text{mole of } O_2 = 28.75 \, \text{moles of } Na \][/tex]
4. Calculate the Mass of [tex]\( Na \)[/tex] Needed:
- Finally, we need to find the mass of [tex]\( Na \)[/tex] required. Using the number of moles and the molar mass of [tex]\( Na \)[/tex]:
[tex]\[ \text{Mass of } Na = 28.75 \, \text{moles} \times 22.99 \, \text{g/mol} \approx 660.9625 \, \text{g} \][/tex]
So, the mass of sodium that will react with [tex]\( 230 \)[/tex] g of oxygen to form sodium oxide is approximately [tex]\( 660.96 \)[/tex] g.
