Need help with these 2 questions ASAP. 1) diiodine pentoxide reacts spectaculary

Question

Need help with these 2 questions ASAP.
1) diiodine pentoxide reacts spectaculary with bromine trifluride to form iodine pentafluride, oxygen gas and liquid bromine. In a particular reaction, a 44.00 g sample of I2O5 reacts with 58.00 g BrF3. What is the mass of the excess reagent? 6I2O5 + 20BrF3 gives 12IF5 + 15O2 + 10 Br2 MM of I2O5 is 333.795 g/mol and the MM of BrF3 is 136.898 g/mol.
2) Balance the following redox equation in acid: Cu3P + Cr2O7^2- gives Cu^2+ + PO4^3- + Cr^3+ Need help with these 2 questions ASAP.
1) diiodine pentoxide reacts spectaculary with bromine trifluride to form iodine pentafluride, oxygen gas and liquid bromine. In a particular reaction, a 44.00 g sample of I2O5 reacts with 58.00 g BrF3. What is the mass of the excess reagent? 6I2O5 + 20BrF3 gives 12IF5 + 15O2 + 10 Br2 MM of I2O5 is 333.795 g/mol and the MM of BrF3 is 136.898 g/mol.
2) Balance the following redox equation in acid: Cu3P + Cr2O7^2- gives Cu^2+ + PO4^3- + Cr^3+ Need help with these 2 questions ASAP.
6I2O5 + 20BrF3 gives 12IF5 + 15O2 + 10 Br2 MM of I2O5 is 333.795 g/mol and the MM of BrF3 is 136.898 g/mol.
2) Balance the following redox equation in acid: Cu3P + Cr2O7^2- gives Cu^2+ + PO4^3- + Cr^3+ MM of I2O5 is 333.795 g/mol and the MM of BrF3 is 136.898 g/mol.
2) Balance the following redox equation in acid: Cu3P + Cr2O7^2- gives Cu^2+ + PO4^3- + Cr^3+
2) Balance the following redox equation in acid: Cu3P + Cr2O7^2- gives Cu^2+ + PO4^3- + Cr^3+

Explanation / Answer

2) Cu+ --> Cu 2+ + 1e-
P 3- --> PO4 3- + 8e-
Cr2O7 2+ + 6e- --> 2Cr 3+
Cu+ --> Cu 2+ + 1e-
P 3- + 4H2O --> PO4 3- + 8e-
Cr2O7 2+ + 6e- --> 2Cr 3+ + 7H2O
Cu+ --> Cu 2+ + 1e-
P 3- + 4H2O --> PO4 3- + 8H+ + 8e-
Cr2O7 2+ + 14 H+ + 6e- --> 2Cr 3+ + 7H2O
Cu+ + Cu 2+ + 1e-
P 3- + 4H2O --> PO4 3- + 8H+ + 8e-
Cr2O7 2+ + 14H+ + 6e- --> 2Cr 3+ + 7H2O + 7H+
2Cu+ --> 2Cu 2+ + 2e-
2P 3- + 8H2O --> 2PO4 3- + 16H+ + 16e-
3Cr2O7 2+ + 42H+ + 18e- --> 3Cr 3+ + 21H2O + 21H+
2Cu+ + 2P 3- + 8H2O + 3Cr2O7 2+ + 42H+ --> 2Cu 2+ + 2PO4 3- + 16H+ + 3Cr 3+ + 21H2O + 21H+
2Cu+ + 2P 3- + 3Cr2O7 2+ + 21H+ --> 2Cu 2+ + 2PO4 3- + 2Cr 3+ + 13H2O

1)20 moles of BrF3 reacts with 6 moles of I2O5

58/136.898 moles react with x

x = 58 * 3/136.898 * 10

= 0.127 moles

So the I2O5 needed is = 0.127 * 333 = 42.291

So the excess reagent is I2O5 And

the excess amount is = 44 - 42.91

=1.09 g

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