Question

1. recall that there are 0.0777 moles $\text{nahco}_3$ reacting in the experiment.

calculate how many moles of $\text{na}$ are produced if reaction #2 occurs.
$2\text{nahco}_3 \longrightarrow 2\text{na} + \text{h}_2 + 2\text{c} + 3\text{o}_2$
i need a hint
mol na
the answer to this question depends on your answers to previous questions.
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Explanation:

Step1: Identify mole ratio

From the balanced equation $2\text{NaHCO}_3
ightarrow 2\text{Na} + \text{H}_2 + 2\text{C} + 3\text{O}_2$, the mole ratio of $\text{NaHCO}_3$ to $\text{Na}$ is $\frac{2\ \text{mol NaHCO}_3}{2\ \text{mol Na}} = 1:1$.

Step2: Calculate moles of Na

Set up proportion using given moles of $\text{NaHCO}_3$:
$\text{Moles of Na} = \text{Moles of NaHCO}_3 \times \frac{2\ \text{mol Na}}{2\ \text{mol NaHCO}_3}$
$\text{Moles of Na} = 0.0777\ \text{mol} \times 1$

Answer:

0.0777 mol