To determine how many moles of [tex]\(\text{Na}_2 \text{O}\)[/tex] will be produced when 2.90 moles of [tex]\(\text{Na}\)[/tex] react completely, we need to look at the stoichiometry of the given chemical reaction:
[tex]\[
4 \text{Na} + \text{O}_2 \rightarrow 2 \text{Na}_2 \text{O}
\][/tex]
From the balanced equation, we can see the molar ratio between sodium ([tex]\(\text{Na}\)[/tex]) and sodium oxide ([tex]\(\text{Na}_2 \text{O}\)[/tex]):
[tex]\[
4 \text{moles of Na} \rightarrow 2 \text{moles of Na}_2 \text{O}
\][/tex]
This ratio can be simplified:
[tex]\[
\frac{2 \text{moles of Na}_2 \text{O}}{4 \text{moles of Na}} = \frac{1 \text{mole of Na}_2 \text{O}}{2 \text{moles of Na}}
\][/tex]
Now, to find out how many moles of [tex]\(\text{Na}_2 \text{O}\)[/tex] are produced from 2.90 moles of [tex]\(\text{Na}\)[/tex], we use the simplified ratio. Let's denote the moles of [tex]\(\text{Na}_2 \text{O}\)[/tex] produced as [tex]\(x\)[/tex].
[tex]\[
\frac{x \text{moles of Na}_2 \text{O}}{2.90 \text{moles of Na}} = \frac{1 \text{mole of Na}_2 \text{O}}{2 \text{moles of Na}}
\][/tex]
Cross-multiplying to solve for [tex]\(x\)[/tex]:
[tex]\[
x \times 2 = 2.90 \times 1
\][/tex]
[tex]\[
2x = 2.90
\][/tex]
[tex]\[
x = \frac{2.90}{2}
\][/tex]
[tex]\[
x = 1.45
\][/tex]
Thus, 2.90 moles of [tex]\(\text{Na}\)[/tex] reacting completely will produce [tex]\(1.45\)[/tex] moles of [tex]\(\text{Na}_2 \text{O}\)[/tex]. Expressed to three significant figures, the answer is:
[tex]\[
\boxed{1.45}
\][/tex]
