Answer to Question #334031 in General Chemistry for nia

Question

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

How many manganese atoms are released if 54.8 moles of Mn3O4 react with excess aluminum.

Answer 1

3 mole of Mn3O4 gives 9 moles of Mn, so 54.8 mole of Mn3O4 gives 54.8/3x9=164.4 mole of Mn

1 mole containse 6.02 1023 atoms.

So 164.4 mole containse 6.02 1023x164.4=990 10²³ atoms of Mn