Chlorine gas was first prepared in 1774 by C. W. Scheele by oxidizing sodium chloride with manganese(IV) oxide. The reaction is $$\mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)$$ Balance this equation.

The unbalanced equation is given as: NaCl(aq) + H2SO4(aq) + MnO2(s) -> Na2SO4(aq) + MnCl2(aq) + H2O(l) + Cl2(g)

On the left side of the equation (reactants), we have: - Sodium (Na): 1 - Chlorine (Cl): 1 - Hydrogen (H): 2 - Sulfur (S): 1 - Manganese (Mn): 1 - Oxygen (O): 6 On the right side of the equation (products), we have: - Sodium (Na): 2 - Chlorine (Cl): 2 - Hydrogen (H): 2 - Sulfur (S): 1 - Manganese (Mn): 1 - Oxygen (O): 5

First, we'll balance the sodium atoms by placing a 2 in front of NaCl: 2NaCl(aq) + H2SO4(aq) + MnO2(s) -> Na2SO4(aq) + MnCl2(aq) + H2O(l) + Cl2(g) Next, we can see that the chlorine atoms are also balanced, but the number of oxygen atoms on the product side is still less than on the reactant side. To fix this issue, we'll add a coefficient of 2 in front of H2O: 2NaCl(aq) + H2SO4(aq) + MnO2(s) -> Na2SO4(aq) + MnCl2(aq) + 2H2O(l) + Cl2(g) Now, counting the atoms again, we have: On the left side of the equation (reactants): - Sodium (Na): 2 - Chlorine (Cl): 2 - Hydrogen (H): 2 - Sulfur (S): 1 - Manganese (Mn): 1 - Oxygen (O): 6 On the right side of the equation (products): - Sodium (Na): 2 - Chlorine (Cl): 2 - Hydrogen (H): 2 - Sulfur (S): 1 - Manganese (Mn): 1 - Oxygen (O): 6 Both sides match, so our balanced chemical equation is: 2NaCl(aq) + H2SO4(aq) + MnO2(s) -> Na2SO4(aq) + MnCl2(aq) + 2H2O(l) + Cl2(g)