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Question
- recall that there are 0.0777 moles $\text{nahco}_3$ reacting.
calculate how many moles of c are produced if reaction #2 occurs when baking soda decomposes.
$2\text{nahco}_3 \
ightarrow 2\text{na} + \text{h}_2 + 2\text{c} + 3\text{o}_2$
i need a hint
mol c
the answer to this question depends on your answers to previous questions.
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Step1: Identify mole ratio
From $2\text{NaHCO}_3
ightarrow 2\text{Na} + \text{H}_2 + 2\text{C} + 3\text{O}_2$, the mole ratio of $\text{NaHCO}_3$ to $\text{C}$ is $\frac{2\ \text{mol NaHCO}_3}{2\ \text{mol C}} = \frac{1\ \text{mol NaHCO}_3}{1\ \text{mol C}}$.
Step2: Calculate moles of C
Set up proportion using given moles of $\text{NaHCO}_3$:
$\text{Moles of C} = 0.0777\ \text{mol NaHCO}_3 \times \frac{1\ \text{mol C}}{1\ \text{mol NaHCO}_3}$
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0.0777 mol C