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Free_Kalibri [48]
3 years ago
5

Balance the following redox equations by the ion-electron method:? A) H2O2 + Fe 2+ ---> Fe 3+ + H2O (in the acidic solution)

C) CN- + MnO4- ---> CNO- +MnO2 (in basic solution) E) S2O2/3- + I2 ---> I- + S4O2/6- (in acidic solution)
Chemistry
1 answer:
Vaselesa [24]3 years ago
3 0
We balance the given reactions above by following the rules in balancing redox reactions in acidic or basic solutions. Balance the atoms aside from the O and H atoms. Then we balance the Os and Hs by adding H2O or H+. Finally, we balance the total charge of the reactant and product by adding e-. We do as follows:

<span>A) H2O2 + Fe 2+ ---> Fe 3+ + H2O (in the acidic solution)
</span><span>   2H+ + </span>H2O2 + Fe 2+ ---> Fe 3+ + 2H2O 
   e- + 2H+ + H2O2 + Fe 2+ ---> Fe 3+ + 2H2O 
<span>
C) CN- + MnO4- ---> CNO- +MnO2 (in basic solution)
</span>     CN- + MnO4- ---> CNO- +MnO2 + H2O
     2H+ + CN- + MnO4- ---> CNO- +MnO2 + H2O
     2OH- + 2H+ + CN- + MnO4- ---> CNO- +MnO2 + H2O + 2OH-
     2H2O + CN- + MnO4- ---> CNO- +MnO2 + H2O + 2OH-
     e- + H2O + CN- + MnO4- ---> CNO- +MnO2 + 2OH-
<span>
E) S2O2/3- + I2 ---> I- + S4O2/6- (in acidic solution)
    2</span>S2O2/3- + I2 ---> 2I- + S4O2/6- 
    4H+ + 2S2O2/3- + I2 ---> 2I- + S4O2/6- + 2H2O
    6e- + 4H+ + 2S2O2/3- + I2 ---> 2I- + S4O2/6- + 2H2O

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The question is incomplete. The complete question is

Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of calcium nitrate. A double displacement reaction occurs. What mass of each of the following substances is present when the reaction stops. A) potassium phosphate remaining B) calcium nitrate g remaining C) calcium phosphate formed D) potassium nitrate g formed

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Explanation:

The equation of the reaction:

2K3PO4(aq) + 3Ca(NO3)2 (aq)-------> 6KNO3(aq) + Ca3(PO4)2(s)

Molar mass of potassium phosphate= 212.27 g/mol

Amount of potassium phosphate= 500/212.27= 2.4 moles

Molar mass of calcium nitrate= 164.088 g/mol

Amount of calcium nitrate= 500/164.088=3.05moles

a) amount of potassium phosphate reacted according to reaction equation= 2 moles

Amount of potassium phosphate remaining= 2.4-2=0.4 moles

Mass of potassium phosphate remaining= 0.4×212.27=84.91g

b) Amount of calcium nitrate reacted according to reaction equation=3

Amount of calcium nitrate remaining=3.05-3= 0.05

Mass of calcium nitrate remaining= 0.05×164.088= 8.20g

c) since calcium nitrate is the limiting reactant, we use to estimate the mass of products formed.

From the reaction equation,

3 moles of calcium nitrate yields 1 mole of calcium phosphate

3.05 moles of calcium nitrate yields 3.05/3 = 1.02 moles of calcium phosphate

Molar Mass of calcium phosphate= 310.18 g/mol

Mass of calcium phosphate produced= 1.02×310.18= 316.4g

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3 moles of calcium nitrate yields 6 moles of potassium nitrate

3.05 moles of calcium nitrate yields 3.05×6/3= 6.1 moles of potassium nitrate

Molar mass of potassium nitrate = 101.1032 g/mol

Mass of potassium nitrate formed= 6.1× 101.1032= 616.73g

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