Consider a 175.7 g sample of the compound manganese(IV) perchlorate.
Using the formula Mn(ClO₄)₄ (MM = 452.74), What quantity in moles of oxygen are in 175.7 g of manganese perchlorate?

Respuesta :

Answer:

6.21 moles O

Explanation:

To find the moles of oxygen, you need to (1) convert grams Mn(ClO₄)₄ to moles Mn(ClO₄)₄ (via molar mass) and then (2) convert moles Mn(ClO₄)₄ to moles O (via mole-to-mole ratio from formula subscripts). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units.

Molar Mass (Mn(ClO₄)₄): 452.74 g/mol

1 Mn(ClO₄)₄ = 1 Mn and 4 Cl and 16 O

175.7 g Mn(ClO₄)₄           1 mole                   16 moles O
---------------------------  x  -------------------  x  ---------------------------  =  6.21 moles O
                                       452.74 g            1 mole Mn(ClO₄)₄