O2 + C10H22 = H2O + CO2

Input interpretation

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oxygen + decane ⟶ water + carbon dioxide

Balanced equation

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$Balance the chemical equation algebraically: + ⟶ + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 + c_4 Set the number of atoms in the reactants equal to the number of atoms in the products for O, C and H: O: | 2 c_1 = c_3 + 2 c_4 C: | 10 c_2 = c_4 H: | 22 c_2 = 2 c_3 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 31/2 c_2 = 1 c_3 = 11 c_4 = 10 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 31 c_2 = 2 c_3 = 22 c_4 = 20 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 31 + 2 ⟶ 22 + 20$

Balance the chemical equation algebraically: + ⟶ + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 + c_4 Set the number of atoms in the reactants equal to the number of atoms in the products for O, C and H: O: | 2 c_1 = c_3 + 2 c_4 C: | 10 c_2 = c_4 H: | 22 c_2 = 2 c_3 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 = 31/2 c_2 = 1 c_3 = 11 c_4 = 10 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 31 c_2 = 2 c_3 = 22 c_4 = 20 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 31 + 2 ⟶ 22 + 20