Balance the following equation: K2CrO4+Na2SO3+HCl→KCl+Na2SO4+CrCl3+H2O 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).

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thomasdecabooter thomasdecabooter · Answer 1 · 2019-12-03T22:41:27+01:00

Answer:

The coefficients found are: 2,3,10,4,3,2,5

The sum of the coefficients is: 2+3+10 + 4 + 3 + 2 + 5 = 29

Explanation:

K2CrO4+Na2SO3+HCl→KCl+Na2SO4+CrCl3+H2O

On the left side we 2 times K , on the right side we have 1 time K

So we have to multiply KCl by 2

K2CrO4+Na2SO3+HCl→2KCl+Na2SO4+CrCl3+H2O

On the left side wehave 1 Cl, on the right side we have 5 Cl

So the left side we have to multiply by 5

K2CrO4+Na2SO3+5 HCl→2KCl+Na2SO4+CrCl3+H2O

On the left side we have 5 times H, on the right side, we have 2 times H

So on the left side the HCl we should multiply by 10 (instead of 5)

On the right side, we should multiply H2O by 5

K2CrO4+Na2SO3+ 10HCl→2KCl+Na2SO4+CrCl3+ 5H2O

So on the left side we have 10 times Cl, on the right side KCl and CrCL3 should be doubled

K2CrO4+Na2SO3+ 10HCl→4 KCl+Na2SO4+ 2CrCl3+ 5H2O

On the right side we have 4 times K, and 2 times Cr. This mean on the left side K2CrO4 should be multiplied by 2

2K2CrO4+Na2SO3+ 10HCl→4 KCl+Na2SO4+ 2CrCl3+ 5H2O

On the left side we have 8 times O plus 3x times O

On the right side we have 5 times O plus 4x times O

To equal this we should multiply on the left side Na2SO3 by 3

and Na2SO4 on the right side also by 3

Now the entire equation is balanced

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

The coefficients found are: 2,3,10,4,3,2,5

The sum of the coefficients is: 2+3+10 + 4 + 3 + 2 + 5 = 29