Question

Fill in the coefficients that will balance the following reaction:

Answer 1

Answer:

a0=1

a1=6

a2=2

a3=3

Explanation:

We have to start with the reaction:

$Cr_2(SO_4)_3~+~AgNO_3~->~Cr(NO_3)_3~+~Ag_2SO_4$

We have to start with "$Cr$". 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:

$Cr_2(SO_4)_3~+~AgNO_3~->~2Cr(NO_3)_3~+~Ag_2SO_4$

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

$Cr_2(SO_4)_3~+~6AgNO_3~->~2Cr(NO_3)_3~+~Ag_2SO_4$

Then we can balance "$Ag$". 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:

$Cr_2(SO_4)_3~+~6AgNO_3~->~2Cr(NO_3)_3~+~3Ag_2SO_4$

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

Answer 2

Answer

Explanation:

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