The volume of Carbon Monoxide gas at STP reacts to produce 759 g of Iron is 458 L.
If this is the Given: [tex] Fe_{2} O_{3} + CO[/tex] >>> [tex] Fe+ CO_{2}[/tex]
Then the first step is to balance it:
[tex] 2 Fe_{2} O_{3} + 6 CO[/tex] >>> [tex] 4 Fe+ 6 CO_{2}[/tex]
Afterwards, we convert 759 g of Iron to moles of Carbon Monoxide (CO).
[tex]759 g Fe * \frac{1 mol Fe}{55.65g Fe} * \frac{6 mol CO}{4 mol Fe} = 20.45822102 mol CO[/tex]
Lastly, we find the volume of carbon monoxide gas at STP, where one mole of gas occupies 22.4 Liters.
[tex]20.45822102 mol CO * \frac{22.4 L of CO}{1 mol CO} = 458.2641509 L of CO[/tex]
Since there are only 3 significant figures, so we round the answer to 458 L.
