Question

If 14.5 g of MnO4- (permanganate) react with manganese (II) hydroxide how many grams of manganese (IV) oxide will be produced? The other products are hydroxide ion and water.

Answer 1

Answer:

$m_{MnO_2}=21.2gMnO_2$

Explanation:

Hello,

In this case, given the balanced reaction:

$2MnO_4^-+2Mn(OH)_2\rightarrow 4MnO_2+2OH^-+H_2O$

We can see a 2:4 mole ration between permanganate ion (118.9 g/mol) and manganese (IV) oxide (86.9 g/mol), that is why the resulting mas of this last one turns out:

$m_{MnO_2}=14.5gMnO_4^-*\frac{1molMnO_4^-}{118.9gMnO_4^-}*\frac{4MnO_2}{2molMnO_4^-} *\frac{86.9gMnO_2}{1molMnO_2} \\\\m_{MnO_2}=21.2gMnO_2$

Best regards.