QQuestionAnatomy and Physiology
QuestionAnatomy and Physiology
2. Assume the reaction is:
2 \mathrm{NaHCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{~g})
What mass of $\mathrm{NaHCO}_{3}$ (s) must have been present at the beginning of the reaction?
# Hide Hint
Hint: Your goal is to do the steps you'd did in the previous section, but in reverse.
You know the mass of the $\mathrm{Na}_{2} \mathrm{CO}_{3}$ product: 3.78 g
The molar mass of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ is $105.99 \mathrm{~g} / \mathrm{mol}$
The molar mass of $\mathrm{NaHCO}_{3}$ is $84.01 \mathrm{~g} / \mathrm{mol}$
How can you use this information to find the mass of the initial amount of $\mathrm{NaHCO}_{3}$ ?
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Answer
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Step 1: Find the number of moles of $\text{Na}_2\text{CO}_3$ produced.
Step 278 \ \cancel{\text{g}} \times \frac{1 \ \text{mol}}{105.99 \ \cancel{\text{g}}} = 0.0357 \ \text{mol} \ \text{of} \ \text{Na}_2\text{CO}_3
Final Answer
6.02 g of $\text{NaHCO}_3$ must have been present at the beginning of the reaction.
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