Answer : The amount of sodium oxide produced is, 74.40 grams
Solution : Given,
Mass of sodium = 55.3 g
Mass of \(O_2\)= 64.3 g
Molar mass of sodium = 23 g/mole
Molar mass of \(O_2\) = 32 g/mole
Molar mass of \(Na_2O\) = 61.9 g/mole
First we have to calculate the moles of sodium and oxygen.
Moles of Na = \(\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= \frac{55.3g}{23g/mole}=2.404moles\)
Moles of \(O_2\) = \(\frac{\text{ given mass of }O_2}{\text{ molar mass of }O_2}= \frac{64.3g}{32g/mole}=2.009moles\)
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction will be,
\(4Na+O_2\rightarrow 2Na_2O\)
From the balanced reaction we conclude that
4 moles of sodium react with 1 mole of oxygen
2.404 moles of sodium react with \(\frac{2.404}{4}=0.601\) moles of oxygen
Excess moles of oxygen = 2.009 - 0.601 = 1.408 moles
That means sodium is a limiting reagent and oxygen is an excess reagent.
Now we have to calculate the moles of sodium oxide.
From the reaction we conclude that,
4 moles of sodium react to give 2 moles of sodium oxide
2.404 moles of sodium react to give \(\frac{2}{4}\times 2.404=1.202\) moles of sodium oxide
Now we have to calculate the mass of sodium oxide.
\(\text{Mass of }Na_2O=\text{Moles of }Na_2O\times \text{Molar mass of }Na_2O\)
\(\text{Mass of }Na_2O=(1.202moles)\times (61.9g/mole)=74.40g\)
Therefore, the amount of sodium oxide produced is, 74.40 grams