To find out how many moles of [tex]\( \text{Na}_2\text{O} \)[/tex] will be produced from the reaction of 2.90 moles of [tex]\( \text{Na} \)[/tex], we can use the stoichiometry of the balanced chemical equation:
[tex]\[ 4 \text{Na} + \text{O}_2 \rightarrow 2 \text{Na}_2\text{O} \][/tex]
From the balanced equation, we can see that 4 moles of [tex]\( \text{Na} \)[/tex] produce 2 moles of [tex]\( \text{Na}_2\text{O} \)[/tex]. This gives us a molar ratio of:
[tex]\[ 4 \text{Na} : 2 \text{Na}_2\text{O} \][/tex]
We can simplify this ratio to:
[tex]\[ 2 \text{Na} : 1 \text{Na}_2\text{O} \][/tex]
This means 2 moles of [tex]\( \text{Na} \)[/tex] produce 1 mole of [tex]\( \text{Na}_2\text{O} \)[/tex].
Given that we have 2.90 moles of [tex]\( \text{Na} \)[/tex], we can use the ratio to find the moles of [tex]\( \text{Na}_2\text{O} \)[/tex]:
[tex]\[ \frac{2.90 \text{ moles Na}}{2} = 1.45 \text{ moles Na}_2\text{O} \][/tex]
Therefore, there will be [tex]\( 1.45 \)[/tex] moles of [tex]\( \text{Na}_2\text{O} \)[/tex] produced, rounded to three significant figures.
There will be [tex]\( 1.45 \)[/tex] moles of [tex]\( \text{Na}_2\text{O} \)[/tex].
