Question
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?
How many milligrams of NaBiO3 are needed to oxidize the manganese in 34.2 mg of manganese(II) sulfate?
Answers
Devron
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
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