Question

How many grams of Na2O are produced when 70.7 g of Na reacts? 4Na(s)+O2(g)→2Na2O(s)

Answer 1

Answer:

95.3 grams.

Explanation:

Relative atomic mass data from a modern periodic table:

• Na: 22.990;
• O: 15.999.

How many moles of Na are consumed?

$M(\mathrm{Na}) = \rm 22.990\; g\cdot mol^{-1}$.

$\displaystyle n(\mathrm{Na}) = \frac{m(\mathrm{Na})}{M(\mathrm{Na})} = \rm \frac{70.7\; g}{22.990\;g\cdot mol^{-1}}=3.07525\;mol$.

How many moles of $\mathrm{Na_2 O}$ formula units will be produced?

Consider the ratio between the coefficient of $\mathrm{Na_2 O}$ and that of $\mathrm{Na}$ in the equation:

$\displaystyle \frac{n(\mathrm{Na_2 O})}{n(\mathrm{Na})} = \frac{2}{4} = \frac{1}{2}$.

As a result,

$\displaystyle n(\mathrm{Na_2 O}) = n(\mathrm{Na}) \cdot \frac{n(\mathrm{Na_2 O})}{n(\mathrm{Na})} = \rm \frac{1}{2}\times 3.07525\;mol = 1.53763\; mol$.

What will be the mass of that many $\mathrm{Na_2 O}$?

Formula mass of $\mathrm{Na_2 O}$:

$M(\mathrm{Na_2 O}) = \rm 2\times 22.990 + 15.999 = 61.979\; g\cdot mol^{-1}$.

$m(\mathrm{Na_2 O}) = n(\mathrm{Na_2 O})\cdot M(\mathrm{Na_2 O}) = \rm 1.53763\; mol\times 61.979\; g\cdot mol^{-1} \approx 95.3\; g$.

Answer 2

Answer: 95,3 g. would be produced, if the mass of Na = 70,7 g.