Question

How many atoms of manganese

are in 250.0 g of braunite,

Mn3SiO6? The molar mass of

braunite is 288.91 g/mol.

? ] × 10[?]

x

atoms Mn

Answer 1

1.56x10^23 atoms of Mn

Step-by-step explanation:

250 grams of braunite (Mn3SiO6) is:

(250 g)/(288.91 g/mole) = 0.865 moles of Mn3SiO6

One mole of Mn3SiO6 would contain 3 moles of Mn, since it appears 3 times in the compound. Breaking apart 1 mole of Mn3SiO6 into individual atoms of each element would procude 3 moles of Mn atoms (and 1 mole of Si and 6 moles of O atoms),

Since we have 0.865 moles of Mn3SiO6,

we should have:

(0.865 moles of Mn3SiO6)*((3 moles of Mn)/(1 mole Mn3SiO6))

this is equal to 2.596 moles of Mn atoms

Since 1 mole is 6.02x10^23 particles of anything, we can write:

(2.596 moles of Mn)*(6.02x10^23 particles Mn atoms/mole) =

1.56x10^23 atoms of Mn