Balance the following equation: K2CrO4+Na2SO3+HCl+KCl+Na2SO4+CrCl3+H2O Enter the

Question

Balance the following equation: K2CrO4+Na2SO3+HCl+KCl+Na2SO4+CrCl3+H2O Enter the coefficients for each compound seperated by commas, in the order in which they appear in the equation.

Explanation / Answer

K2CrO4 + Na2SO3 + HCl -----> KCl + Na2SO4 + CrCl3 + H2O Cr goes from +6 in K2CrO4 to +3 in CrCl3; Cr gains 3e- (reduced) S goes from +4 in Na2SO3 to +6 in Na2SO4; S loses 2e- (oxidized) To balance the electrons, take the Cr compounds times 2 and the S compounds times 3 to get 6 electrons gained and lost. This gives: 2K2CrO4 + 3Na2SO3 + HCl -----> KCl + 3Na2SO4 + 2CrCl3 + H2O Next, balance the oxygens by inspection (17 oxygens on each side). 2K2CrO4 + 3Na2SO3 + HCl -----> KCl + 3Na2SO4 + 2CrCl3 + 5H2O Balance the hydrogens (10 hydrogens on each side) 2K2CrO4 + 3Na2SO3 + 10HCl -----> KCl + 3Na2SO4 + 2CrCl3 + 5H2O Balance the chlorines (10 Cl on each side) 2K2CrO4 + 3Na2SO3 + 10HCl -----> 4KCl + 3Na2SO4 + 2CrCl3 + 5H2O It is now balanced. There are 4 potassiums on each side.

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