Oxidation of MnCl2 by NaBiO3: 5.0g Yields How Many Grams?

[sveioen]

Sodium bismuthate (NaBiO3) oxidizes MnCl2 to MnO4 and is consequently reduced to Bi(OH)3. How many grams of MnCl2 will be oxidized by 5.0 g of NaBiO3?

Could anyone please help me with this? I don't have a clue where to begin.

 

[chemisttree]

Begin by writing out a balanced equation for this redox system.

 

[Blueshift5]

I can provide a response to this question by using the principles of stoichiometry and the balanced chemical equation for the reaction.

First, we need to determine the molar ratios between NaBiO3 and MnCl2 in the balanced equation. From the equation, we can see that for every 1 mole of NaBiO3, 1 mole of MnCl2 is oxidized to form 1 mole of MnO4. This means that the molar ratio between NaBiO3 and MnCl2 is 1:1.

Next, we need to calculate the moles of NaBiO3 present in 5.0 g of NaBiO3. To do this, we need to know the molar mass of NaBiO3, which is 261.99 g/mol. Using the formula n = m/M, where n is the number of moles, m is the mass, and M is the molar mass, we can calculate that 5.0 g of NaBiO3 contains 0.0191 moles (5.0 g / 261.99 g/mol = 0.0191 mol).

Since the molar ratio between NaBiO3 and MnCl2 is 1:1, we can infer that 0.0191 moles of NaBiO3 will oxidize 0.0191 moles of MnCl2. To convert this to grams, we need to multiply the number of moles by the molar mass of MnCl2, which is 125.84 g/mol. This means that 0.0191 moles of MnCl2 will be oxidized by 2.4 grams (0.0191 mol x 125.84 g/mol = 2.4 g) of NaBiO3.

Therefore, 5.0 g of NaBiO3 will oxidize approximately 2.4 grams of MnCl2.