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Fill in the coefficients that will balance the following reaction:
a0Cr2(SO4)3 + a1AgNO3 -> a2Cr(NO3)3 + a3Ag2SO4

Respuesta :

AmieB

Answer

Explanation:

can you please put it in boxes so i can find out the answer

Answer:

a0=1

a1=6

a2=2

a3=3

Explanation:

We have to start with the reaction:

[tex]Cr_2(SO_4)_3~+~AgNO_3~->~Cr(NO_3)_3~+~Ag_2SO_4[/tex]

We have to start with "[tex]Cr[/tex]". If we want to balance Cr we have to obtain 2 on both sides. Therefore we have to add a "2" on the right side, so:

[tex]Cr_2(SO_4)_3~+~AgNO_3~->~2Cr(NO_3)_3~+~Ag_2SO_4[/tex]

Then we can balance "[tex]NO_3[/tex]" as a whole. We have 6 on the right, we have to have the same amount on the left side, so:

[tex]Cr_2(SO_4)_3~+~6AgNO_3~->~2Cr(NO_3)_3~+~Ag_2SO_4[/tex]

Then we can balance "[tex]Ag[/tex]". We have 6 on the left, so we have to have the same amount on the right if we already have 2 we have to put a · in front (on the right side) to obtain in total 6, so:

[tex]Cr_2(SO_4)_3~+~6AgNO_3~->~2Cr(NO_3)_3~+~3Ag_2SO_4[/tex]

Finally, when we add this last number the [tex]SO_4[/tex] we will have 3 on both sides. Therefore the reaction is already balanced.

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