Redox balancing problem

AsO3 + NO3 --> H3AsO4 + N2O3

it will be: AsO3 --> H3AsO4 and NO3 --> N2O3

I saw in the answer sheet that the oxidation part of the reaction became
5H20 + AsO3 --> 2H3AsO4 + 4H + 4e.
I was wondering why there's a 2 in the H3AsO4 and why isn't it H + 5H20 + AsO3 --> 2H3AsO4? Thank you so much.

The oxide is As2O3, not AsO3

As2O3 --> H3AsO4

in balancing redox reactions, first balance elements other than O and H

As2O3 --> 2H3AsO4

balance O first by adding H2O:

5H2O + As2O3 -------> 2H3AsO4 (there are 8 O on the right--you need 5 more on the left to balance)

then add H+ ions to balance H:

5H2O + As2O3 ----> 2H3AsO4 + 4H+

then and e- (electrons) to balance charge

5H2O + As2O3 ----> 2H3AsO4 + 4H+ + 4e-

Is it AsO3^-3 as it cannot be AsO3 where As has an oxidation no. of +6 because the O.N. in product is +5 because the other substance NO3- is also getting reduced. Both cannot be reduced.

AsO3^-3 + H2O ---> AsO4^-3 + 2H+ + 2e
2 No3- + 6H+ + 4e ----> N2O3 + 3 H2O
Multiplying 1st with 2 and adding to cancel e
2NO3- + 2H+ + 2AsO3^-3 ------> N2O3 + H2O + 2 AsO4^-3 or
2HNO3 + 2 H3AsO3 ----> N2O3 + H2O + 2 H3AsO4