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H2SO4 + K2Cr2O7 + C2H5OH = H2O + K2SO4 + Cr2(SO4)3 + C2H4O

Input interpretation

H_2SO_4 sulfuric acid + K_2Cr_2O_7 potassium dichromate + CH_3CH_2OH ethanol ⟶ H_2O water + K_2SO_4 potassium sulfate + Cr_2(SO_4)_3 chromium sulfate + C_2H_4O ethylene oxide
H_2SO_4 sulfuric acid + K_2Cr_2O_7 potassium dichromate + CH_3CH_2OH ethanol ⟶ H_2O water + K_2SO_4 potassium sulfate + Cr_2(SO_4)_3 chromium sulfate + C_2H_4O ethylene oxide

Balanced equation

Balance the chemical equation algebraically: H_2SO_4 + K_2Cr_2O_7 + CH_3CH_2OH ⟶ H_2O + K_2SO_4 + Cr_2(SO_4)_3 + C_2H_4O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2SO_4 + c_2 K_2Cr_2O_7 + c_3 CH_3CH_2OH ⟶ c_4 H_2O + c_5 K_2SO_4 + c_6 Cr_2(SO_4)_3 + c_7 C_2H_4O Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, S, Cr, K and C: H: | 2 c_1 + 6 c_3 = 2 c_4 + 4 c_7 O: | 4 c_1 + 7 c_2 + c_3 = c_4 + 4 c_5 + 12 c_6 + c_7 S: | c_1 = c_5 + 3 c_6 Cr: | 2 c_2 = 2 c_6 K: | 2 c_2 = 2 c_5 C: | 2 c_3 = 2 c_7 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 1 c_3 = 3 c_4 = 7 c_5 = 1 c_6 = 1 c_7 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 4 H_2SO_4 + K_2Cr_2O_7 + 3 CH_3CH_2OH ⟶ 7 H_2O + K_2SO_4 + Cr_2(SO_4)_3 + 3 C_2H_4O
Balance the chemical equation algebraically: H_2SO_4 + K_2Cr_2O_7 + CH_3CH_2OH ⟶ H_2O + K_2SO_4 + Cr_2(SO_4)_3 + C_2H_4O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2SO_4 + c_2 K_2Cr_2O_7 + c_3 CH_3CH_2OH ⟶ c_4 H_2O + c_5 K_2SO_4 + c_6 Cr_2(SO_4)_3 + c_7 C_2H_4O Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, S, Cr, K and C: H: | 2 c_1 + 6 c_3 = 2 c_4 + 4 c_7 O: | 4 c_1 + 7 c_2 + c_3 = c_4 + 4 c_5 + 12 c_6 + c_7 S: | c_1 = c_5 + 3 c_6 Cr: | 2 c_2 = 2 c_6 K: | 2 c_2 = 2 c_5 C: | 2 c_3 = 2 c_7 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 1 c_3 = 3 c_4 = 7 c_5 = 1 c_6 = 1 c_7 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 H_2SO_4 + K_2Cr_2O_7 + 3 CH_3CH_2OH ⟶ 7 H_2O + K_2SO_4 + Cr_2(SO_4)_3 + 3 C_2H_4O