Answer to Question #143972 in General Chemistry for liam donohue

Question

Using the following balanced chemical equation, answer the following questions based on starting mass of 346.33g Fe2O3

Fe2O3+3CO-2FE+3CO2

The MM (g/mol) of Fe203?
Moles (mol) of Fe203?
Moles (mol) of CO2?
MM (g/mol) of CO2?
Final mass (g) of CO2?

Answer 1

346.33g Fe2O3

Fe2O3+3CO-2FE+3CO2

The MM (g/mol) of Fe203

Molar mass = 159.69g

Moles (mol) of Fe203

159.69g = 1 mole

346.33g Fe2O3 = 346.33/159.69

= 2.1688 moles

Moles (mol) of CO2

(2.1688 moles )(3)

= 6.5063moles

MM (g/mol) of CO2

44.01 g/mol = 1mole

? = 6.5063

(44.01)(6.5063)= 286.34g

Final mass (g) of CO2?

44.01)(6.5063)= 286.34g

159.69g = 1 mole

346.33g Fe2O3 = 346.33/159.69

= 2.1688 moles