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H2SO4 + K2Cr2O7 + CH3CH2OH = H2O + K2SO4 + Cr2(SO4)3 + CH3COOH

Input interpretation

sulfuric acid + potassium dichromate + ethanol ⟶ water + potassium sulfate + chromium sulfate + acetic acid
sulfuric acid + potassium dichromate + ethanol ⟶ water + potassium sulfate + chromium sulfate + acetic acid

Balanced equation

Balance the chemical equation algebraically:  + + ⟶ + + +  Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 + c_3 ⟶ c_4 + c_5 + c_6 + c_7  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 + 2 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 = 3/2 c_4 = 11/2 c_5 = 1 c_6 = 1 c_7 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 8 c_2 = 2 c_3 = 3 c_4 = 11 c_5 = 2 c_6 = 2 c_7 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 8 + 2 + 3 ⟶ 11 + 2 + 2 + 3
Balance the chemical equation algebraically: + + ⟶ + + + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 + c_3 ⟶ c_4 + c_5 + c_6 + c_7 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 + 2 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 = 3/2 c_4 = 11/2 c_5 = 1 c_6 = 1 c_7 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 8 c_2 = 2 c_3 = 3 c_4 = 11 c_5 = 2 c_6 = 2 c_7 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 8 + 2 + 3 ⟶ 11 + 2 + 2 + 3

Structures

 + + ⟶ + + +
+ + ⟶ + + +

Names

sulfuric acid + potassium dichromate + ethanol ⟶ water + potassium sulfate + chromium sulfate + acetic acid
sulfuric acid + potassium dichromate + ethanol ⟶ water + potassium sulfate + chromium sulfate + acetic acid

Chemical names and formulas

 | sulfuric acid | potassium dichromate | ethanol | water | potassium sulfate | chromium sulfate | acetic acid Hill formula | H_2O_4S | Cr_2K_2O_7 | C_2H_6O | H_2O | K_2O_4S | Cr_2O_12S_3 | C_2H_4O_2 name | sulfuric acid | potassium dichromate | ethanol | water | potassium sulfate | chromium sulfate | acetic acid IUPAC name | sulfuric acid | dipotassium oxido-(oxido-dioxochromio)oxy-dioxochromium | ethanol | water | dipotassium sulfate | chromium(+3) cation trisulfate | acetic acid
| sulfuric acid | potassium dichromate | ethanol | water | potassium sulfate | chromium sulfate | acetic acid Hill formula | H_2O_4S | Cr_2K_2O_7 | C_2H_6O | H_2O | K_2O_4S | Cr_2O_12S_3 | C_2H_4O_2 name | sulfuric acid | potassium dichromate | ethanol | water | potassium sulfate | chromium sulfate | acetic acid IUPAC name | sulfuric acid | dipotassium oxido-(oxido-dioxochromio)oxy-dioxochromium | ethanol | water | dipotassium sulfate | chromium(+3) cation trisulfate | acetic acid