Question

1. construct the mole ratio needed to calculate how many moles of $na_2o$ would you expect to be produced if reaction #3 occurs.$2\text{nahco}_3 \

ightarrow \text{na}_2\text{o} + 2\text{co}_2 + \text{h}_2\text{o}$recall that there are 0.0777 moles $\text{nahco}_3$ reacting in the experiment.i need a hint$\frac{\text{a units}}{\text{ }} \times \frac{\text{c units}}{\text{e units}} = ? \text{ mol } \text{na}_2\text{o}$choices:0.0777mole $\text{o}_2$mole $\text{na}_2\text{o}$mole $\text{nahco}_3$12

Explanation:

Step1: Identify mole ratio from reaction

From the balanced equation $2\text{NaHCO}_3
ightarrow \text{Na}_2\text{O} + 2\text{CO}_2 + \text{H}_2\text{O}$, the mole ratio of $\text{Na}_2\text{O}$ to $\text{NaHCO}_3$ is $\frac{1\ \text{mol Na}_2\text{O}}{2\ \text{mol NaHCO}_3}$.

Step2: Set up mole conversion

Multiply given moles of $\text{NaHCO}_3$ by the mole ratio:
$\text{Moles of Na}_2\text{O} = 0.0777\ \text{mol NaHCO}_3 \times \frac{1\ \text{mol Na}_2\text{O}}{2\ \text{mol NaHCO}_3}$

Step3: Calculate final value

Cancel units and compute:
$\text{Moles of Na}_2\text{O} = \frac{0.0777}{2} = 0.03885$

Filling the blanks:
A: $0.0777$
C: $1\ \text{mol Na}_2\text{O}$
E: $2\ \text{mol NaHCO}_3$

Answer:

Filled calculation: $0.0777\ \text{mol NaHCO}_3 \times \frac{1\ \text{mol Na}_2\text{O}}{2\ \text{mol NaHCO}_3} = 0.03885\ \text{mol Na}_2\text{O}$
Final moles of $\text{Na}_2\text{O}$: $0.0389$ (rounded to 3 significant figures)