H2SO4 + K2Cr2O7 + K2S = H2O + K2SO4 + S + Cr2(SO4)3

Input interpretation

Image
Text

sulfuric acid + potassium dichromate + K2S ⟶ water + potassium sulfate + mixed sulfur + chromium sulfate

Balanced equation

Image
Text

$Balance the chemical equation algebraically: + + K2S ⟶ + + + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 + c_3 K2S ⟶ c_4 + c_5 + c_6 + c_7 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, S, Cr and K: H: | 2 c_1 = 2 c_4 O: | 4 c_1 + 7 c_2 = c_4 + 4 c_5 + 12 c_7 S: | c_1 + c_3 = c_5 + c_6 + 3 c_7 Cr: | 2 c_2 = 2 c_7 K: | 2 c_2 + 2 c_3 = 2 c_5 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_2 = 1 and solve the system of equations for the remaining coefficients: c_2 = 1 c_3 = (3 c_1)/4 - 9/4 c_4 = c_1 c_5 = (3 c_1)/4 - 5/4 c_6 = c_1 - 4 c_7 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_2 = 2 c_3 = (3 c_1)/4 - 9/2 c_4 = c_1 c_5 = (3 c_1)/4 - 5/2 c_6 = c_1 - 8 c_7 = 2 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 10 and solve for the remaining coefficients: c_1 = 10 c_2 = 2 c_3 = 3 c_4 = 10 c_5 = 5 c_6 = 2 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 + 2 + 3 K2S ⟶ 10 + 5 + 2 + 2$

Balance the chemical equation algebraically: + + K2S ⟶ + + + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 + c_3 K2S ⟶ 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 and K: H: | 2 c_1 = 2 c_4 O: | 4 c_1 + 7 c_2 = c_4 + 4 c_5 + 12 c_7 S: | c_1 + c_3 = c_5 + c_6 + 3 c_7 Cr: | 2 c_2 = 2 c_7 K: | 2 c_2 + 2 c_3 = 2 c_5 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_2 = 1 c_3 = (3 c_1)/4 - 9/4 c_4 = c_1 c_5 = (3 c_1)/4 - 5/4 c_6 = c_1 - 4 c_7 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_2 = 2 c_3 = (3 c_1)/4 - 9/2 c_4 = c_1 c_5 = (3 c_1)/4 - 5/2 c_6 = c_1 - 8 c_7 = 2 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 10 and solve for the remaining coefficients: c_1 = 10 c_2 = 2 c_3 = 3 c_4 = 10 c_5 = 5 c_6 = 2 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 + 2 + 3 K2S ⟶ 10 + 5 + 2 + 2