Answer: a) [tex]Fe_2O_3+3CO\rightarrow 2Fe+3CO_2[/tex]
b) 33749996 grams
Explanation:
a) The balanced chemical reactions will be :
[tex]3Fe_2O_3+CO\rightarrow CO_2+2Fe_3O_4[/tex]
[tex]2Fe_3O_4+2CO\rightarrow 6FeO+2CO_2[/tex]
[tex]6FeO+6CO\rightarrow 6Fe+6CO_2[/tex]
Overall equation : [tex]3Fe_2O_3+9CO\rightarrow 6Fe+9CO_2[/tex]
Overall balanced equation : [tex]Fe_2O_3+3CO\rightarrow 2Fe+3CO_2[/tex]
b) Amount of iron = 45 metric ton = 45000 kg = 45000000g
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe=\frac{45000000g}{56g/mol}=803571moles[/tex]
[tex]Fe_2O_3+3CO\rightarrow 2Fe+3CO_2[/tex]
According to stoichiometry :
2 moles of [tex]Fe[/tex] are produced from= 3 moles of [tex]CO[/tex]
Thus 803571 moles of [tex]Fe[/tex] are produced from=[tex]\frac{3}{2}\times 803571=1205357moles[/tex] of [tex]CO[/tex]
Mass of [tex]CO=moles\times {\text {Molar mass}}=1205357moles\times 28g/mol=33749996g[/tex]
Thus 33749996 g of carbon monoxide are required to form 45.0 metric tons of iron from ferric oxide