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

`m_(MnO_2)=21.2gMnO_2`

Step-by-step 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^-*(1molMnO_4^-)/(118.9gMnO_4^-)*(4MnO_2)/(2molMnO_4^-) *(86.9gMnO_2)/(1molMnO_2) \\\\m_(MnO_2)=21.2gMnO_2`

Best regards.