Balance the following equation: K_2CrO_4 + Na_2SO_3 + HCl rightarrow KCl + Na_2S

Question

Balance the following equation: K_2CrO_4 + Na_2SO_3 + HCl rightarrow KCl + Na_2SO_4 + CrCl_3 + H_2 O Generally coefficients of 1 are omitted from balanced chemical equations. When entering your answer, include coefficients of 1 as required for grading purposes. Enter the coefficients for each compound, separated by commas, in the order in which they appear in the equation (e.g., 1, 2, 3, 4, 5, 6, 7). In the process of oxidizing I^- to I_2. SO_4^2- is reduced to SO_2. How many moles of SO_2 are produced in the formation of one mole of I_2? Express your answer numerically in moles.

Explanation / Answer

1.   Half reaction method

CrO4^2-(aq)====>Cr^3+(aq)
add water to the right to balance O
CrO4^2-(aq)====>Cr^3+(aq) + 4H2O(l)
add H+ to the left to balance H
CrO4^2-(aq) + 8H+(aq)=====>Cr^3+(aq) + 4H2O(l)
add e- to the left to balance charge
CrO4^2-(aq) + 8H+(aq) + 3e-=====>Cr^3+(aq) + 4H2O(l)

SO3^2-(aq)====>SO4^2-(aq)
add H2O to the left to balance O
SO3^2-(aq) + H2O(l)====>SO4^2-(aq)
add H+ to the right to balance H
SO3^2-(aq) + H2O(l)=====>SO4^2-(aq) + 2H+(aq)
add e- to the right to balance charge
SO3^2-(aq) + H2O(l)=====>SO4^2-(aq) + 2H+(aq) + 2e-

balanced half reactions are:

CrO4^2-(aq) + 8H+(aq) + 3e-=====>Cr^3+(aq) + 4H2O(l)
SO3^2-(aq) + H2O(l)=====>SO4^2-(aq) + 2H+(aq) + 2e-
multiply the first one by 2 and the second one by 3 then combine to get rid of e-.

2CrO4^2-(aq) + 10H+(aq) + 3SO3^2-(aq)====>2Cr^3+(aq) + 5H2O(l) + 3SO4^2-(aq)

That's the net ionic equation. Now add back spectator ions, clean it up and you're done.

2K2CrO4 + 3Na2SO3 + 10HCl====>4KCl + 3Na2SO4 + 2CrCl3 + 5H2O

2. 2I I2 + 2e

SO42 + 4H+ + 2e SO2 + 2H2O

2 I + SO42- + 4 H+ I2 + SO2 + 2 H2O

(1 mol I2) x (1 mol SO2 / 1 mol I2) = 1 mol SO2

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