Solution A is 0.2400 M Ca(NO3)2(aq). Solution A is diluted by a factor of 3 to form Solution B. As2O3(s) reacts with NO3¯(aq) to produce NO(g) and H3ASO4(aq) When As2O3(s) was added to 27.50 mL of Solution B, 1.710 × 10²1 molecules of NO(g) were obtained from the reaction in 88.75% yield. (a) If As2O3(s) was the limiting reactant, calculate the mass of As2O3(s) which reacted with NO3 (aq). (b) Determine the number of moles of NO3 in the 27.50 mL of Solution B which remained unreacted at the end of the reaction.
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The reaction taking place is
3As2O3 + 4NO3- + 7H2O + 4H+ → 6H3AsO4 + 4NO
Balancing explanation :
As in As2O3 is in As3+ state and in H3AsO4 its in As5+ state and N in NO3- is in 5+ state and N in NO is in 2+ state
Hence oxidation of As and reduction of N is happening
Hence the half reactions will be
Reduction : NO3- + 3 e- ---------> NO
Since we have 3 O in LHS, hence adding 2 H2O in RHS to balance it
=> NO3- + 3 e- ---------> NO + 2 H2O
Now we have 4 H in RHS, hence adding 4H+ in LHS to balance it
=> NO3- + 4H+ + 3 e- ---------> NO + 2 H2O
Oxidation : As2O3 --------> 2 H3AsO4 + 4 e-
Since we have 8 O in RHS, hence making it 8 in LHS
=> As2O3 + 5 H2O --------> 2 H3AsO4 + 4 e-
Now we have 10 H in LHS while 6 in RHS
hence making it 10 in RHS
=> As2O3 + 5 H2O --------> 2 H3AsO4 + 4H+ + 4 e-
Now since we have both oxidation and reduction balanced reactions
Hence multiplying 3 to the oxidation half and 4 to the reduction half and adding both cancelling 12 electron to get net balanced reaction as
4 NO3- + 4 H+ + 3 As2O3 + 7 H2O --------> 6 H3AsO4 + 4 NO
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