To determine the mass of sodium chloride ([tex]\( \text{NaCl} \)[/tex]) formed when 7.5 grams of sodium carbonate ([tex]\( \text{Na}_2\text{CO}_3 \)[/tex]) reacts with hydrochloric acid ([tex]\( \text{HCl} \)[/tex]), we must follow a systematic approach. Let's break it down step-by-step.
1. Determine moles of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex]:
[tex]\[
\text{Moles of } \text{Na}_2\text{CO}_3 = \frac{\text{mass of } \text{Na}_2\text{CO}_3 \ \text{(g)}}{\text{molar mass of } \text{Na}_2\text{CO}_3 \ \text{(g/mol)}}
\][/tex]
Given:
- Mass of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] is [tex]\( 7.5 \)[/tex] grams.
- Molar mass of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] is [tex]\( 105.99 \)[/tex] g/mol.
[tex]\[
\text{Moles of } \text{Na}_2\text{CO}_3 = \frac{7.5 \text{ g}}{105.99 \text{ g/mol}} \approx 0.0708 \text{ moles}
\][/tex]
2. Using the balanced chemical equation:
The balanced chemical equation is:
[tex]\[
\text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2
\][/tex]
According to the equation, 1 mole of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] produces 2 moles of [tex]\( \text{NaCl} \)[/tex].
[tex]\[
\text{Moles of } \text{NaCl} = 2 \times \text{Moles of } \text{Na}_2\text{CO}_3
\][/tex]
[tex]\[
\text{Moles of } \text{NaCl} = 2 \times 0.0708 \approx 0.1415 \text{ moles}
\][/tex]
3. Calculate the mass of [tex]\( \text{NaCl} \)[/tex] produced:
[tex]\[
\text{Mass of } \text{NaCl} = \text{moles of } \text{NaCl} \times \text{molar mass of } \text{NaCl}
\][/tex]
Given the molar mass of [tex]\( \text{NaCl} \)[/tex] is [tex]\( 58.44 \)[/tex] g/mol:
[tex]\[
\text{Mass of } \text{NaCl} = 0.1415 \text{ moles} \times 58.44 \text{ g/mol} \approx 8.28 \text{ grams}
\][/tex]
So, the mass of sodium chloride ([tex]\( \text{NaCl} \)[/tex]) formed is approximately [tex]\( 8.28 \)[/tex] grams, which corresponds to the correct number of significant figures based on the given data.
