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Common Polyatomic Ions

\begin{tabular}{|c|c|c|c|}
\hline
Chemical Name & Chemical Formula & Chemical Name & Chemical Formula \\
\hline
acetate & [tex]$C_2H_3O_2^{-}$[/tex] & nitrite & [tex]$NO_2^{-}$[/tex] \\
\hline
carbonate & [tex]$CO_3^{2-}$[/tex] & ammonium & [tex]$NH_4^{+}$[/tex] \\
\hline
hypocarbonite & [tex]$CO^{2-}$[/tex] & cyanide & [tex]$CN^{-}$[/tex] \\
\hline
\begin{tabular}{l}
hydrogen carbonate \\
(bicarbonate)
\end{tabular} & [tex]$HCO_3^{-}$[/tex] & hydroxide & [tex]$OH^{-}$[/tex] \\
\hline
chlorite & [tex]$ClO_2^{-}$[/tex] & peroxide & [tex]$O_2^{2-}$[/tex] \\
\hline
\end{tabular}

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Type the correct answer in the box. Express your answer to three significant figures.

This balanced equation shows the reaction of sodium hydroxide and sulfuric acid:

[tex]\[ 2 NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2 H_2O \][/tex]

In a laboratory experiment, a student mixes 355 grams of sulfuric acid with an excess of sodium hydroxide. What is the theoretical mass of sodium sulfate produced? Refer to the periodic table and the polyatomic ion resource.

The theoretical mass of sodium sulfate is [tex]$\square$[/tex] grams.

Sagot :

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To solve this problem, we need to follow a systematic approach based on stoichiometry. Here's the detailed, step-by-step solution:

1. Write the balanced chemical equation:
[tex]\[ 2 \, \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \, \text{H}_2\text{O} \][/tex]
This equation shows that 1 mole of sulfuric acid ([tex]\(\text{H}_2\text{SO}_4\)[/tex]) reacts with 2 moles of sodium hydroxide ([tex]\(\text{NaOH}\)[/tex]) to produce 1 mole of sodium sulfate ([tex]\(\text{Na}_2\text{SO}_4\)[/tex]) and 2 moles of water ([tex]\(\text{H}_2\text{O}\)[/tex]).

2. Determine the molar masses:
- Molar mass of sulfuric acid ([tex]\(\text{H}_2\text{SO}_4\)[/tex]):
[tex]\[ \text{H}_2\text{SO}_4 = 2 \times 1\, \text{g/mol (H)} + 32\, \text{g/mol (S)} + 4 \times 16\, \text{g/mol (O)} = 2 + 32 + 64 = 98\, \text{g/mol} \][/tex]
- Molar mass of sodium sulfate ([tex]\(\text{Na}_2\text{SO}_4\)[/tex]):
[tex]\[ \text{Na}_2\text{SO}_4 = 2 \times 23\, \text{g/mol (Na)} + 32\, \text{g/mol (S)} + 4 \times 16\, \text{g/mol (O)} = 46 + 32 + 64 = 142\, \text{g/mol} \][/tex]

3. Calculate the moles of sulfuric acid used:
Given the mass of sulfuric acid is 355 grams:
[tex]\[ \text{Moles of } \text{H}_2\text{SO}_4 = \frac{355\, \text{g}}{98\, \text{g/mol}} = 3.622 \, \text{moles} \, (\text{to three significant figures}) \][/tex]

4. Determine the moles of sodium sulfate produced:
From the balanced equation, 1 mole of [tex]\(\text{H}_2\text{SO}_4\)[/tex] produces 1 mole of [tex]\(\text{Na}_2\text{SO}_4\)[/tex]. Thus, 3.622 moles of [tex]\(\text{H}_2\text{SO}_4\)[/tex] will produce:
[tex]\[ \text{Moles of } \text{Na}_2\text{SO}_4 = 3.622 \, \text{moles} \][/tex]

5. Calculate the mass of sodium sulfate produced:
[tex]\[ \text{Mass of } \text{Na}_2\text{SO}_4 = \text{Moles of } \text{Na}_2\text{SO}_4 \times \text{Molar mass of } \text{Na}_2\text{SO}_4 \][/tex]
[tex]\[ \text{Mass of } \text{Na}_2\text{SO}_4 = 3.622 \, \text{moles} \times 142\, \text{g/mol} = 514\, \text{g} \, (\text{to three significant figures}) \][/tex]

Therefore, the theoretical mass of sodium sulfate produced is [tex]\(\boxed{514}\)[/tex] grams.
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