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 g. would be produced, if the mass of Na = 70,7 g.

Answer 2

95.3 grams.

Step-by-step 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 (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})} = (2)/(4) = (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 (1)/(2)* 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* 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* 61.979\; g\cdot mol^(-1) \approx 95.3\; g`.