Math

QuestionChlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq)\mathrm{HCl}(\mathrm{aq}), as described by the chemical equation MnO2( s)+4HCl(aq)MnCl2(aq)+2H2O(l)+Cl2( g)\mathrm{MnO}_{2}(\mathrm{~s})+4 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{MnCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{Cl}_{2}(\mathrm{~g})
How much MnO2( s)\mathrm{MnO}_{2}(\mathrm{~s}) should be added to excess HCl(aq)\mathrm{HCl}(\mathrm{aq}) to obtain 225 mLCl2( g)225 \mathrm{~mL} \mathrm{Cl}_{2}(\mathrm{~g}) at 25C25^{\circ} \mathrm{C} and 725 Torr? mass of MnO2\mathrm{MnO}_{2} : \square g

Studdy Solution
Calculate the mass of MnO2\mathrm{MnO}_{2}:
Using the molar mass of MnO2\mathrm{MnO}_{2} (86.94g/mol86.94 \, \text{g/mol}): mass=nMnO2×molar mass\text{mass} = n_{\mathrm{MnO}_{2}} \times \text{molar mass} mass=0.00915mol×86.94g/mol\text{mass} = 0.00915 \, \text{mol} \times 86.94 \, \text{g/mol}
Calculate the mass: mass0.796g\text{mass} \approx 0.796 \, \text{g}
The mass of MnO2\mathrm{MnO}_{2} needed is:
0.796g\boxed{0.796 \, \text{g}}

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