JEE MAIN - Chemistry (2022 - 29th June Morning Shift - No. 1)
Production of iron in blast furnace follows the following equation
Fe3O4(s) + 4CO(g) $$\to$$ 3Fe(l) + 4CO2(g)
when 4.640 kg of Fe3O4 and 2.520 kg of CO are allowed to react then the amount of iron (in g) produced is :
[Given : Molar Atomic mass (g mol$$-$$1) : Fe = 56, Molar Atomic mass (g mol$$-$$1) : O = 16, Molar Atomic mass (g mol$$-$$1) : C = 12]
Explanation
Given,
Mass of $$F{e_3}{O_4}$$ = 4.640 kg = 4640 gm
Molar mass of $$F{e_3}{O_4}$$ = 56 $$\times$$ 3 + 16 $$\times$$ 4 = 232 g
$$\therefore$$ Moles of $$F{e_3}{O_4} = {{4640} \over {232}} = 20$$
Also, given
Mass of CO = 2.520 kg = 2520 gm
Molar mass of CO = 12 + 16 = 28 gm
$$\therefore$$ Molar of $$CO = {{2520} \over {28}} = 90$$
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Here Fe3O4 is limiting reagent as to find limiting reagent, divide the given moles of reactants with their respective stoichiometric coefficient and reactant for which this ratio is minimum will be limiting reagent
For $$F{e_3}{O_4},{{moles} \over {stoichiometric\,coefficient}} = {{20} \over 1}$$
For $$CO,{{moles} \over {stoichiometric\,coefficient}} = {{90} \over 4} = 22.5$$
$$\therefore$$ Fe3O4 is limiting reagent.
Now produced Fe = 20 $$\times$$ 3 = 60 mol
$$\therefore$$ Weight of Fe = 60 $$\times$$ 56 = 3360 g
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