Questions: Here's the balanced reaction: 2 NaHCO3(s) -> 2 Na(s) + H2(g) + 2 C(s) + 3 O2(g) The mole-to-mole ratio from the balanced equation is: 2 moles NaHCO3 = 2 moles C Since the mole ratio is 2:2, the math set-up looks like this: 0.064 mole NaHCO3 x (2 mole C / 2 mole NaHCO3) = ? mole C

Here's the balanced reaction:
2 NaHCO3(s) -> 2 Na(s) + H2(g) + 2 C(s) + 3 O2(g)

The mole-to-mole ratio from the balanced equation is:

2 moles NaHCO3 = 2 moles C
Since the mole ratio is 2:2, the math set-up looks like this:
0.064 mole NaHCO3 x (2 mole C / 2 mole NaHCO3) = ? mole C

Transcript text: Here's the balanced reaction: \[ \begin{array}{l} 2 \mathrm{NaHCO}_{3}(\mathrm{~s}) \rightarrow 2 \mathrm{Na}(\mathrm{~s})+\mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{C}(\mathrm{~s})+ \\ 3 \mathrm{O}_{2}(\mathrm{~g}) \end{array} \] The mole-to-mole ratio from the balanced equation is: 2 moles $\mathrm{NaHCO}_{3}=2$ moles C Since the mole ratio is $2: 2$, the math set-up looks like this: 0.064 mole $\mathrm{NaHCO}_{3} \times \frac{2 \mathrm{~mole} \mathrm{C}^{2}}{2 \mathrm{moleNaHCO}_{3}}=$ ? mole C

Solution

Solution Steps

Step 1: Identify the Mole Ratio

From the balanced chemical equation: \[ 2 \mathrm{NaHCO}_{3} \rightarrow 2 \mathrm{Na} + \mathrm{H}_{2} + 2 \mathrm{C} + 3 \mathrm{O}_{2} \] The mole-to-mole ratio between $\mathrm{NaHCO}_{3}$ and $\mathrm{C}$ is $2:2$.

Step 2: Set Up the Conversion

Using the mole ratio, we can set up the conversion for the given amount of $\mathrm{NaHCO}_{3}$: \[ 0.064 \text{ mole } \mathrm{NaHCO}_{3} \times \frac{2 \text{ mole } \mathrm{C}}{2 \text{ mole } \mathrm{NaHCO}_{3}} \]

Step 3: Perform the Calculation

Simplify the conversion factor: \[ 0.064 \text{ mole } \mathrm{NaHCO}_{3} \times 1 = 0.064 \text{ mole } \mathrm{C} \]