Balance the redox reaction below in acid solution: BiO_3^- + Fe^2+right arrow Bi
ID: 942335 • Letter: B
Question
Balance the redox reaction below in acid solution: BiO_3^- + Fe^2+right arrow Bi^3+ + Fe^3+ 50.00ml of BiO_3^- is titrated with 26.43ml 0.02982M Fe^2+. What is the molar concentration of BiO_3^-? Flow many mg of NaBiO_3were in the 50.00ml sample? What volume ofO_2 (in ml) is produced when 25.00 g of 3.00%(m/m) H_2O_2 is completely reduced by the addition of excess MnO_4, at 20.0degreeC and 748 mmHg. What is the volume strength of this solution (assuming d_H2O2 = 1.00g/ml)? H_2O_2 + MnO_4^- right arrow O_2 + Mn^2+Explanation / Answer
Answer – 2) We are given the [Fe2+] = 0.02982 M , volume = 26.43 mL
Volume of BiO3- = 50.00 mL.
Moles of Fe2+ = 0.02980 M * 0.02643 L
= 0.000788 moles
From the balanced reaction equation
2 moles of Fe2+ = 1 moles of BiO3-
So, 0.000788 moles of Fe2+ = ?
= 0.000394 moles of BiO3-
So, [BiO3-] = 0.000394 moles / 0.050 L
= 0.00788 M
We know, 1 mole of BiO3- = 1 moles of NaBiO3 = 0.000394 moles
So, mass of NaBiO3 = 0.000394 moles * 279.968 g/mol
= 0.110 g
= 110 mg
3) Given, 25.00 g of 3.00 % (m/m) H2O2 , T = 20.0oC , P = 748 mmHg
Density of H2O2 = 1.00 g/mL
Balanced reaction –
5H2O2 + 2MnO4- + 6H+------> 5 O2 + 2 Mn2++ 8H2O
Mass of H2O2 = 25 g *3.00 % / 100 %
= 0.75 g
Moles of H2O2 = 0.75 g / 34.016 g.mol-1
= 0.0220 moles
From the balanced equation –
5 moles of H2O2 = 5 moles of O2
So, 0.0220 mole of H2O2 = ?
= 0.0220 moles of O2
At 20oC the water vapor pressure = 17.5 mm Hg
So, P of O2 = 748 mm Hg – 17.5 mm Hg
= 730.5 mm Hg
= 0.961 atm
using the ideal gas law volume of O2
PV = nRT
So, V = nRT/P
= 0.0220 moles * 0.0821 L.atm/mol.K*293 K / 0.961 atm
= 0.552 L
= 552 mL
Volume of H2O2 = 0.75 g / 1.00 g/mL
= 0.75 mL
So, volume strength = 552 mL / 0.75 mL
= 736 mL O2/ mL H2O2
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