Answer:
95.3 grams.
Explanation:
Relative atomic mass data from a modern periodic table:
How many moles of Na are consumed?
[tex]M(\mathrm{Na}) = \rm 22.990\; g\cdot mol^{-1}[/tex].
[tex]\displaystyle n(\mathrm{Na}) = \frac{m(\mathrm{Na})}{M(\mathrm{Na})} = \rm \frac{70.7\; g}{22.990\;g\cdot mol^{-1}}=3.07525\;mol[/tex].
How many moles of [tex]\mathrm{Na_2 O}[/tex] formula units will be produced?
Consider the ratio between the coefficient of [tex]\mathrm{Na_2 O}[/tex] and that of [tex]\mathrm{Na}[/tex] in the equation:
[tex]\displaystyle \frac{n(\mathrm{Na_2 O})}{n(\mathrm{Na})} = \frac{2}{4} = \frac{1}{2}[/tex].
As a result,
[tex]\displaystyle n(\mathrm{Na_2 O}) = n(\mathrm{Na}) \cdot \frac{n(\mathrm{Na_2 O})}{n(\mathrm{Na})} = \rm \frac{1}{2}\times 3.07525\;mol = 1.53763\; mol[/tex].
What will be the mass of that many [tex]\mathrm{Na_2 O}[/tex]?
Formula mass of [tex]\mathrm{Na_2 O}[/tex]:
[tex]M(\mathrm{Na_2 O}) = \rm 2\times 22.990 + 15.999 = 61.979\; g\cdot mol^{-1}[/tex].
[tex]m(\mathrm{Na_2 O}) = n(\mathrm{Na_2 O})\cdot M(\mathrm{Na_2 O}) = \rm 1.53763\; mol\times 61.979\; g\cdot mol^{-1} \approx 95.3\; g[/tex].
