Question

1)

A) If I want to make 225.0 g of Fe, how many grams of C will I need for my experiment?

B) If I want to make 225g of Fe, how many grams of Fe₂O₃ will I need for my experiment?

2) There are 3.00 moles of oxygen gas and 6.00 moles of nitrogen gas in a 10.0 L container at 300K.

A) What is the total pressure (in atm) of the gas mixture?

B) What is the partial pressure (in atm) of O₂ in the gas mixture?

C) What is the partial pressure (at atm) of N₂ in the gas mixture?

Answer 1

1)

A)

Fe2O3 + 3 C ------------------> 3CO + 2 Fe

                 36 g                                       111.7 g

                x g                                            225 g

x = 36 x 225 / 111.7 = 72.5g

mass of C needed = 72.5 g

B)

Fe2O3    + 3 C ------------------> 3CO + 2 Fe

159.69g                                                   111.7 g

                                                                   225 g

mass of Fe2O3 = 159.69 x 225 / 111.7

                         = 321.7 g needed

2) There are 3.00 moles of oxygen gas and 6.00 moles of nitrogen gas in a 10.0 L container at 300K.

total moles = 3 + 6 = 9

P V = n R T

P x 10 = 9 x 0.0821 x 300

P = 22.17 g

A) total pressure (in atm) = 22.2 atm

B) What is the partial pressure (in atm) of O₂ in the gas mixture?

partial pressure of O2 = 7.4 atm

C) What is the partial pressure (at atm) of N₂ in the gas mixture

partial pressure of N2 = 14.8 atm