Answer:
H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-
Explanation:
Reduction half reaction
2H_2O + CrO_4^2- + 3e -> CrO_2^- + 4OH^-
Oxidation half reaction
2OH^- + SO_3^2- -> SO_4^2- + H_2O + 2e
Balanced overall equation
H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-
Answer: m= 3.15x10-3 g NaHCO3
Explanation: To find the mass of NaHCO3 we will use the relationship between moles and molar mass. The molar mass of NaHCO3 is 84 g.
3.75x10-5 moles NaHCO3 x 84 g NaHCO3 / 1 mole NaHCO3
= 3.15x10-3 g NaHCO3