SuperDude23
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At a very high temperature, manganese is isolated from its ore, manganomanganic oxide, via the following balanced equation:

3 Mn3O4(s) + 8 Al(s)  4 Al2O3(s) + 9 Mn(s)

How many grams of Mn3O4 are required to produce 10.0g of Mn?

You will need to calculate the molar mass of Mn3O4


10.0 grams


0.20 grams


5.0 grams


13.9 grams

At a very high temperature manganese is isolated from its ore manganomanganic oxide via the following balanced equation 3 Mn3O4s 8 Als 4 Al2O3s 9 Mns How many g class=
At a very high temperature manganese is isolated from its ore manganomanganic oxide via the following balanced equation 3 Mn3O4s 8 Als 4 Al2O3s 9 Mns How many g class=

Respuesta :

First, we need to calculate the molar mass of Mn3O4. Molar mass of Mn = 54.94 g/mol, O = 16.00 g/mol. Therefore, Molar mass of Mn3O4 = (3 * 54.94) + (4 * 16.00) = 228.82 g/mol

Now, we can use stoichiometry to find the amount of Mn3O4 required:

10.0 g Mn * (1 mol Mn3O4 / 9 mol Mn) * (228.82 g Mn3O4 / 1 mol Mn3O4) = 25.42 g Mn3O4

So, the correct answer is not listed, the calculated value is 25.42 grams of Mn3O4.

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