To determine how many grams of [tex]\( \text{Cl}_2(g) \)[/tex] could be prepared by the reaction of [tex]\( 100.0 \)[/tex] grams of [tex]\( \text{HCl} (aq) \)[/tex] with excess [tex]\( \text{MnO}_2(s) \)[/tex], we need to follow several steps systematically. Here’s a detailed step-by-step solution:
1. Write down the balanced chemical equation:
[tex]\[
\text{MnO}_2(s) + 4 \text{HCl}(aq) \rightarrow \text{MnCl}_2(aq) + 2 \text{H}_2\text{O}(l) + \text{Cl}_2(g)
\][/tex]
According to the balanced equation, 4 moles of [tex]\(\text{HCl}\)[/tex] produce 1 mole of [tex]\(\text{Cl}_2\)[/tex].
2. Determine the molar mass of the substances involved:
- Molar mass of [tex]\(\text{HCl} = 1.01 + 35.45 = 36.46 \ \text{g/mol}\)[/tex]
- Molar mass of [tex]\(\text{Cl}_2 = 2 \times 35.45 = 70.90 \ \text{g/mol}\)[/tex]
3. Calculate the moles of [tex]\(\text{HCl}\)[/tex] used:
Given mass of [tex]\(\text{HCl}\)[/tex] is [tex]\( 100.0 \ \text{g} \)[/tex].
[tex]\[
\text{Moles of HCl} = \frac{\text{Mass of HCl}}{\text{Molar mass of HCl}} = \frac{100.0 \ \text{g}}{36.46 \ \text{g/mol}} = 2.7427317608337902 \ \text{moles}
\][/tex]
4. Determine the moles of [tex]\(\text{Cl}_2\)[/tex] produced:
From the balanced equation, [tex]\( 4 \)[/tex] moles of [tex]\(\text{HCl}\)[/tex] produce [tex]\( 1 \)[/tex] mole of [tex]\(\text{Cl}_2\)[/tex].
[tex]\[
\text{Moles of Cl}_2 = \frac{\text{Moles of HCl}}{4} = \frac{2.7427317608337902}{4} = 0.6856829402084476 \ \text{moles}
\][/tex]
5. Calculate the mass of [tex]\(\text{Cl}_2\)[/tex] produced:
[tex]\[
\text{Mass of Cl}_2 = \text{Moles of Cl}_2 \times \text{Molar mass of Cl}_2 = 0.6856829402084476 \ \text{moles} \times 70.90 \ \text{g/mol} = 48.61492046077893 \ \text{g}
\][/tex]
6. Conclusion:
The mass of [tex]\( \text{Cl}_2(g) \)[/tex] that could be prepared by the reaction of [tex]\(\text{100.0 g HCl} (aq)\)[/tex] with excess [tex]\(\text{MnO}_2(s)\)[/tex] is [tex]\( 48.61 \ \text{g} \)[/tex]. (Rounded to two decimal places for simplicity.)
