- Calculate the molar mass of \(\text{Na}_2\text{CO}_3\) as \(105.99\text{ g/mol}\) and the molar mass of \(\text{NaCl}\) as \(58.44\text{ g/mol}\).
- Convert \(7.5\text{ g}\) of \(\text{Na}_2\text{CO}_3\) into approximately \(0.0708\text{ moles}\).
- Apply the stoichiometric ratio of \(2\text{ NaCl} : 1\text{ Na}_2\text{CO}_3\) to find the moles of \(\text{NaCl}\) produced, approximately \(0.1415\text{ moles}\).
- Calculate the mass of \(\text{NaCl}\) as approximately \(8.3\text{ g}\) when rounded to two significant figures.
- The final mass of sodium chloride is \(\boxed{8.3\text{ g}}\).
Explanation
- Problem Analysis and Data Preparation
The problem asks us to determine the mass of sodium chloride (\(\text{NaCl}\)) produced from the reaction of \(7.5\text{ g}\) of sodium carbonate (\(\text{Na}_2\text{CO}_3\)) with hydrochloric acid (\(\text{HCl}\)). The balanced chemical equation for this reaction is:
\[\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\]
The stoichiometric ratio indicates that \(1\) mole of \(\text{Na}_2\text{CO}_3\) reacts with \(2\) moles of \(\text{HCl}\) to produce \(2\) moles of \(\text{NaCl}\). The input mass of \(7.5\text{ g}\) has two significant figures, so our final result should be reported to two significant figures. We will use the molar masses of the elements: \(\text{Na} = 22.99\text{ g/mol}\), \(\text{C} = 12.01\text{ g/mol}\), \(\text{O} = 16.00\text{ g/mol}\), and \(\text{Cl} = 35.45\text{ g/mol}\).
- Calculating Moles of Reactant
First, we calculate the molar mass of sodium carbonate (\(\text{Na}_2\text{CO}_3\)):
\[M(\text{Na}_2\text{CO}_3) = 2 \times 22.99 + 12.01 + 3 \times 16.00 = 45.98 + 12.01 + 48.00 = 105.99\text{ g/mol}\]
Next, we find the number of moles of \(\text{Na}_2\text{CO}_3\) in \(7.5\text{ g}\):
\[n(\text{Na}_2\text{CO}_3) = \frac{7.5\text{ g}}{105.99\text{ g/mol}} \approx 0.07076\text{ mol}\]
- Calculating Mass of Product and Final Result
From the balanced equation, \(1\text{ mol}\) of \(\text{Na}_2\text{CO}_3\) produces \(2\text{ mol}\) of \(\text{NaCl}\). Therefore, the number of moles of \(\text{NaCl}\) produced is:
\[n(\text{NaCl}) = 2 \times n(\text{Na}_2\text{CO}_3) = 2 \times 0.07076\text{ mol} = 0.14152\text{ mol}\]
Now, calculate the molar mass of \(\text{NaCl}\):
\[M(\text{NaCl}) = 22.99 + 35.45 = 58.44\text{ g/mol}\]
Finally, calculate the mass of \(\text{NaCl}\) produced:
\[\text{mass}(\text{NaCl}) = n(\text{NaCl}) \times M(\text{NaCl}) = 0.14152\text{ mol} \times 58.44\text{ g/mol} \approx 8.2706\text{ g}\]
Rounding to two significant figures as per the input data (\(7.5\text{ g}\)), the mass is approximately \(8.3\text{ g}\).
Examples
This type of stoichiometric calculation is crucial in industrial chemistry for determining the amount of raw materials needed for large-scale production. For instance, if a manufacturing facility needs to produce sodium chloride for food processing, they use these calculations to figure out exactly how much sodium carbonate must be reacted with acid to get the target yield, minimizing waste and ensuring efficiency. Understanding stoichiometry allows scientists to predict the outcome of reactions accurately, which is fundamental to chemical manufacturing, environmental testing, and pharmaceutical development.