Iron(III) Oxide (Fe2O3) reacts wth CO to produce iron (Fe) and CO2 by the follow

Question

Iron(III) Oxide (Fe2O3) reacts wth CO to produce iron (Fe) and CO2 by the following equation:

                         Fe2O3(s) + 3CO(g) ----> 2Fe(s) + 3CO2(g)

If one starts with 30.0g Fe2O3(s) and 10.0g CO(g). How many grams of iron can be produced?

2) Pure titanium metal is produced commercially from titanium (IV) Chloride (TiCl4), which is produced via the reaction:

                     TiO2(s) + 2C(s) + 2Cl2(g) --------> TiCl4(l) + 2CO(g)

How much TiCl4 can be made from 2.0 mol TiO2(s), 5.0 mol C(s), and 3.0 mol Cl2(g)?

3) Let's say we're to determine empirical formula. I did that and got AlNa3. But then they ask "if the molecular mass  is 200 amu, what is the molecular formula? Do I simply multiply molar mass of AlNa3 by 2? I dont get it.

Explanation / Answer

1)no.of moles of iron oxide = 30/160 = 0.1875 moles [since mol wt of iron(lll)oxide is 160]

no. of moles of CO = 10/28 = 0.35714

1 mol Fe2O3 reacts with 3 moles of CO

=> 0.1875 moles Fe2O3 requires 0.5625 moles of CO

since CO is present in less amount ,it is the limitting reagent

now product formation is w.r.t CO

since 3 moles CO gives 2 moles Fe

=> 0.35714 moles CO gives 2*0.35714/3 moles Fe = 0.3281 moles Fe

= 0.3281*56 g Fe = 13.3333g Fe can be produced ans...

2) it can be seen seen that Cl2 is the limitting reagent

since 2 moles Cl2 gives 1 mole TiCl4

=> 3 moles Cl2 gives 3/2 =1.5 moles TiCl4 = 190*1.5 g = 285 g of TiCl4 ans...

3) empirical formula mass = 27 + 23*3 = 96

n= molecular mass / empirical formula mass.

=> n = 200 / 96 = 2.0833 ~ 2

=> molecular formula = (empirical formula)*n

=> molecular formula = Al2Na6 ans...

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