The theoretical mass of Na₂O that should remain for Reaction II is approximately 2.79 g, which is closest to the option 2.78 g Na₂O.
To determine the mass of solid product Na₂O that should remain for Reaction II, we need to calculate the theoretical yield of Na₂O from the initial mass of NaHCO₃ and compare it with the given experimental data.
Reaction II:
2NaHCO₃(s) → Na₂O(s) + 2CO₂(g) + H₂O(g)
From the balanced equation, we can see that the molar ratio between NaHCO₃ and Na₂O is 2:1.
This means that for every 2 moles of NaHCO₃ reacted, 1 mole of Na₂O is produced.
First, let's calculate the number of moles of NaHCO₃:
\[ \text{Moles of NaHCO₃} = \frac{\text{Initial mass of NaHCO₃}}{\text{Molar mass of NaHCO₃}} \]
Given that the molar mass of NaHCO₃ is approximately 84 g/mol, we can calculate:
\[ \text{Moles of NaHCO₃} = \frac{7.53 \, \text{g}}{84 \, \text{g/mol}} \approx 0.09 \, \text{mol} \]
Now, since the molar ratio between NaHCO₃ and Na₂O is 2:1, the number of moles of Na₂O produced is half of the moles of NaHCO₃:
\[ \text{Moles of Na₂O} = (1)/(2) * \text{Moles of NaHCO₃} \approx (1)/(2) * 0.09 \, \text{mol} = 0.045 \, \text{mol} \]
Next, we calculate the theoretical mass of Na₂O using its molar mass, which is approximately 62 g/mol:
\[ \text{Theoretical mass of Na₂O} = \text{Moles of Na₂O} * \text{Molar mass of Na₂O} \]\[ \text{Theoretical mass of Na₂O} = 0.045 \, \text{mol} * 62 \, \text{g/mol} \approx 2.79 \, \text{g} \]
Therefore, the theoretical mass of Na₂O that should remain for Reaction II is approximately 2.79 g.
Among the given options, the closest one is 2.78 g Na₂O.
