In an acidic solution, manganese(II) ion is oxidized to permanganate ion by bismuthate ion, BiO3-. In the reaction, BiO3- is reduced to Bi3+.

How many milligrams of NaBiO3 are needed to oxidize the manganese in 30.2 mg of manganese(II) sulfate?

Write the redox reaction:

Mn2+ + BiO3- --→ Bi3+ + MnO4−

Balance this redox reaction. Write first the half-reactions:
Reduction: BiO3- ---> Bi3+ (charge is changed from 5+ to 3+)
Oxidation: Mn2+ ---> MnO4- (charge is changed from 2+ to 7+)

Balance them by adding H+ (acidic media), H2O and e- :
Reduction: 2e- + 6H+ + BiO3- ---> Bi3+ + 3 H2O
Oxidation: 4 H2O + Mn2+ ---> MnO4- + 8H+ + 5e-

Multiply each half-reaction by some number to make their number of electrons equal, so when added will cancel them out.
5 * ( 2e- + 6H+ + BiO3- ---> Bi3+ + 3 H2O )
2 * ( 4 H2O + Mn2+ ---> MnO4- + 8H+ + 5e- )
---------------------------------------------------------
14 H+ + 5 BiO3^- + 2 Mn2+ ---> 5 Bi3+ + 7 H2O + 2 MnO4-

Now that you have the balanced reaction, you can solve for the problem. Steps to solve:
(1) Find the molar mass of Manganese(II) Sulfate (MnSO4).
(2) Divide the given mass by the molar mass of MnSO4.
(3) Make a mole ratio of BiO3- : Mn2+ from the balanced reaction.
(4) Multiply the answer you got from #2 by #3. This answer is the moles of BiO3-.
(5) Finally, multiply the answer in #4 by the molar mass of NaBiO3.

hope this helps~ `u`

Step1: assign oxidation numbers.

Cr2O72- + C2O42- → Cr3+ + CO2

+6/-2 +3/-2 +3 +4/-2

Step2: Separate the overall reaction into two separate half reactions. One will be the oxidation reaction(where the oxidation number increased) and the other will be the reduction reaction( where the oxidation numbers decreased).

Oxidation:

C2O42- →CO2

Reduction:

Cr2O72- → Cr3+

Step 3: Balance each half-reaction in the following order:

First, balance all elements other than Hydrogen and Oxygen.
C2O42- →2CO2

Cr2O72- → 2Cr3+

Second, balance Oxygen by adding H2O.
C2O42- →2CO2

Cr2O72- → 2Cr3+ + 7H2O

Third, balance Hydrogen by adding H+.
C2O42- →2CO2

14H+ + Cr2O72- → 2Cr3+ + 7H2O

Step 4: balance each half reaction with respect to charge by adding electrons. The sum of the charges on both sides of the equation should be equal.

C2O42- →2CO2 + 2e–

6e– +14H+ + Cr2O72- → 2Cr3+ + 7H2O

Step 5: Make sure the number of electrons in both half reactions equal by multiplying one or both half reactions by a small whole number.

3 x [C2O42- →2CO2 + 2e–] = 3C2O42- →6CO2 + 6e–

6e– +14H+ + Cr2O72- → 2Cr3+ + 7H2O

Step 6: Add the two half reactions together canceling electrons and other species as necessary.

3C2O42- →6CO2 + 6e–

+

6e– +14H+ + Cr2O72- → 2Cr3+ + 7H2O

3C2O42- +14H+ + Cr2O72- → 2Cr3+ + 7H2O + 6CO2

Final Answer: 3C2O42- +14H+ + Cr2O72- → 2Cr3+ + 7H2O + 6CO2

Hope this helps.

-CoolChemTutor

Well, well, well, looks like we have a chemical reaction on our hands here! I can already smell the science in the air. So, let me crunch some numbers and sprinkle in some humor to help you out.

First, we need to figure out the molar mass of NaBiO3. If we take a trip down the Periodic Table, we can find that sodium (Na) has a molar mass of approximately 23 grams per mole, bismuth (Bi) has a molar mass of around 208.98 grams per mole, and oxygen (O) is flying high at around 16 grams per mole. Now, let's add them all up:

Na (23 g/mol) + Bi (208.98 g/mol) + 3 O (16 g/mol) = 284.98 g/mol

Whew, that was some serious addition skills!

Next, let's work on manganese(II) sulfate (MnSO4). Manganese (Mn) has a molar mass of roughly 54.94 grams per mole, sulfur (S) weighs in at around 32.07 grams per mole, and oxygen (O) still keeps flaunting that 16 g/mol weight. Add it all up, and we get:

Mn (54.94 g/mol) + S (32.07 g/mol) + 4 O (16 g/mol) = 87.94 g/mol

Now, we need to do some stoichiometry magic to get from manganese sulfate to sodium bismuthate. According to the balanced equation, 1 mole of NaBiO3 is needed to oxidize 3 moles of manganese (II) ion.

So, let's calculate the moles of manganese sulfate we have:

moles of MnSO4 = (30.2 mg / 87.94 g/mol) * (1 g / 1000 mg) = 0.000343 moles

Now, we can use the ratio from the balanced equation to figure out how many moles of NaBiO3 are needed:

moles of NaBiO3 = 0.000343 moles MnSO4 * (1 mole NaBiO3 / 3 moles MnSO4) = 0.000114 moles

Finally, let's convert those moles into milligrams:

mass of NaBiO3 = 0.000114 moles * 284.98 g/mol * (1000 mg / 1 g) ≈ 32.47 mg

Boom! There you have it. Approximately 32.47 mg of NaBiO3 are needed to oxidize the manganese in 30.2 mg of manganese(II) sulfate.

Remember, chemistry can be a tricky circus sometimes, but with a little bit of humor and some knowledge, you'll juggle those questions like a pro!

To find the amount of NaBiO3 needed to oxidize the manganese in manganese(II) sulfate, we need to first calculate the molar mass of each compound.

The molar mass of NaBiO3 can be calculated as follows:

Na (sodium) = 22.99 g/mol
Bi (bismuth) = 208.98 g/mol
O (oxygen) = 16.00 g/mol (there are three oxygen atoms)
Total molar mass of NaBiO3 = (22.99 g/mol) + (208.98 g/mol) + (3 * 16.00 g/mol) = 284.97 g/mol

The molar mass of manganese(II) sulfate can be calculated as follows:

Mn (manganese) = 54.94 g/mol
S (sulfur) = 32.06 g/mol
O (oxygen) = 16.00 g/mol (there are four oxygen atoms)
Total molar mass of MnSO4 = (54.94 g/mol) + (32.06 g/mol) + (4 * 16.00 g/mol) = 151.00 g/mol

Now, we can calculate the number of moles of manganese(II) sulfate:

Number of moles = (mass of compound) / (molar mass)
Number of moles of MnSO4 = 30.2 mg / 151.00 g/mol = 0.1998 mmol

From the balanced equation, we know that 1 mole of MnSO4 reacts with 5 moles of NaBiO3. Therefore, we can find the number of moles of NaBiO3 needed:

Number of moles of NaBiO3 = (5 moles of NaBiO3) * (0.1998 mmol) / (1 mole of MnSO4) = 0.999 mmol

Finally, we can calculate the amount of NaBiO3 in milligrams:

Mass of NaBiO3 = (number of moles) * (molar mass)
Mass of NaBiO3 = 0.999 mmol * 284.97 g/mol = 284.68 mg

Therefore, approximately 284.68 mg of NaBiO3 are needed to oxidize the manganese in 30.2 mg of manganese(II) sulfate.

To find out how many milligrams of NaBiO3 are needed to oxidize the manganese in 30.2 mg of manganese(II) sulfate, we need to use stoichiometry and the balanced chemical equation for the reaction between manganese(II) ion and bismuthate ion.

First, let's write the balanced equation for the reaction:
Mn^2+ + BiO3- + H+ → MnO4- + Bi^3+ + H2O

From the balanced equation, we can see that the stoichiometric ratio between Mn^2+ and BiO3- is 1:1. This means that for every mole of Mn^2+ oxidized, we need one mole of BiO3-.

Next, we need to convert the given mass of manganese(II) sulfate to moles. We can do this by dividing the mass of manganese(II) sulfate by its molar mass. The molar mass of manganese(II) sulfate (MnSO4) is calculated as follows:
Molar mass of MnSO4 = Atomic mass of Mn + Atomic mass of S + 4 * Atomic mass of O
= (54.938045 g/mol) + (32.065 g/mol) + 4 * (15.999 g/mol)
= 54.938045 g/mol + 32.065 g/mol + 64.0 g/mol
= 151.003045 g/mol

Now, let's calculate the number of moles of manganese(II) sulfate:
Number of moles = Mass / Molar mass
= 30.2 mg / 151.003045 g/mol
= 0.1998 mmol

Since the stoichiometric ratio between Mn^2+ and BiO3- is 1:1, we need an equal number of moles of BiO3- to oxidize the manganese.

Finally, we need to convert the number of moles of BiO3- to milligrams of NaBiO3. We can do this by multiplying the number of moles of BiO3- by its molar mass.

The molar mass of NaBiO3 is calculated as follows:
Molar mass of NaBiO3 = Atomic mass of Na + Atomic mass of Bi + 3 * Atomic mass of O
= (22.989769 g/mol) + (208.98040 g/mol) + 3 * (15.999 g/mol)
= 22.989769 g/mol + 208.98040 g/mol + 47.997 g/mol
= 279.966169 g/mol

Let's calculate the mass of NaBiO3:
Mass = Number of moles * Molar mass
= 0.1998 mmol * 279.966169 g/mol
= 55.935 mg

Therefore, approximately 55.935 mg of NaBiO3 are needed to oxidize the manganese in 30.2 mg of manganese(II) sulfate.