- #1
lorka150
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This is the full question:
Suppose you roast 2.00kg of Fe3S4. The equation:
4Fe3S4 + 25 O2 --> 6Fe2O3 + 16SO2.
a) How many kg of Fe2O3 will be produced?
b) How many kg of SO2 will be produced?
c) How many litres of air at 25 degrees C are required for roasting? Assuming air has 23% O2 by mass, and density of 25C is 1.2 g/L.
The molar masses are Fe3S4 = 295.8 g/mol, Fe2O3 = 159.69 g/mol and SO2 is 64.06 g/mol.
I have done a and b.
a) = 1.62kg
b) = 1.73kg.
I have NO IDEA how to do C! Please guide!
Suppose you roast 2.00kg of Fe3S4. The equation:
4Fe3S4 + 25 O2 --> 6Fe2O3 + 16SO2.
a) How many kg of Fe2O3 will be produced?
b) How many kg of SO2 will be produced?
c) How many litres of air at 25 degrees C are required for roasting? Assuming air has 23% O2 by mass, and density of 25C is 1.2 g/L.
The molar masses are Fe3S4 = 295.8 g/mol, Fe2O3 = 159.69 g/mol and SO2 is 64.06 g/mol.
I have done a and b.
a) = 1.62kg
b) = 1.73kg.
I have NO IDEA how to do C! Please guide!