Potable water (drinking water) should not have manganese concentrations in excess of 0.05 mg/mL. If the manganese concentration is greater than 0.1 mg/mL, it imparts a foul taste to the water and discolors laundry and porcelain surfaces. 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 34.2 mg of manganese(II) sulfate?
14 H^+ + 2 Mn^2+ + 5 NaBiO3 --->5 Bi^3+ + 2 MnO4^- + 5 Na^+ + 7 H2O
Here is the the balanced reaction (Oxidation and Reduction)
34.2MgSO4=0.0342g of MgSO4
0.0342g of MgSO4*(1 mole/120.366 g)= moles of MgSO4
The equation shows that 2 MnSO4 moles=5 moles of NaBiO3
So,
moles of MgSO4*(5 moles of NaBiO3/ 2 moles of MnSO4)=moles of NaBiO3
moles of NaBiO3*(279.97 g/mol)=NaBiO3 in g
NaBiO3 in g*(10^3mg/g)= answer in mg
****Answer contains 3-significant figures
To determine the amount of NaBiO3 needed to oxidize the manganese in 34.2 mg of manganese(II) sulfate, we need to use stoichiometry and the balanced chemical equation to calculate the molar ratios of the reactants.
The balanced chemical equation for the reaction is as follows:
2 MnSO4 + 5 NaBiO3 + 8 H2SO4 → 5 Na2SO4 + 2 MnSO4 + 5 Bi2(SO4)3 + 4 H2O
According to the balanced equation, 5 moles of NaBiO3 are required to oxidize 2 moles of MnSO4.
First, we need to convert the mass of MnSO4 to moles using its molar mass.
The molar mass of MnSO4 is calculated as follows:
Molar mass of Mn = 54.94 g/mol
Molar mass of S = 32.07 g/mol
4 times the molar mass of O = 16.00 g/mol
Total molar mass of MnSO4 = (54.94 + 32.07 + 4 * 16.00) g/mol = 151.00 g/mol
To convert the mass of MnSO4 to moles, we use the equation:
moles = mass / molar mass
moles of MnSO4 = 34.2 mg / 151.00 g/mol = 0.2264 mol
Now we can calculate the moles of NaBiO3 needed using the mole ratio from the balanced equation.
moles of NaBiO3 = (5 moles NaBiO3 / 2 moles MnSO4) * 0.2264 mol MnSO4
moles of NaBiO3 = 0.1132 mol NaBiO3
Finally, we can convert the moles of NaBiO3 to mass using its molar mass.
The molar mass of NaBiO3 is calculated as follows:
Molar mass of Na = 22.99 g/mol
Molar mass of Bi = 208.98 g/mol
3 times the molar mass of O = 16.00 g/mol
Total molar mass of NaBiO3 = (22.99 + 208.98 + 3 * 16.00) g/mol = 284.97 g/mol
mass of NaBiO3 = moles of NaBiO3 * molar mass of NaBiO3
mass of NaBiO3 = 0.1132 mol * 284.97 g/mol = 32.26 g
Therefore, 32.26 grams of NaBiO3 are needed to oxidize the manganese in 34.2 mg of manganese(II) sulfate.