Use the reaction:?

Question

Use the reaction: () are subscripts
2Fe(2)O3 + 3C -> 4Fe + 3CO(2)


Using 1000g C, how many grams of Fe can be produced?

What is the theoretical yield of Fe in a reaction with 7.00g Fe(2)O(3) and 1.25g C?

What amount of the reactant in excess will remain at the end of the reaction? Give answer in grams.

Answer 1

3 moles of carbon will yield 4 moles of Fe; 1 mole of C yields
(4/3) mole Fe. The atomic mass of C is 12.01, so 1000g = 1000/12.01 = 83.26 moles. (4/3) times this is 111.02; that is the moles of Fe. The atomic mass of Fe is 55.85, so 111.02 moles of Fe is 111.02*55.85 = 6200 g Fe

1.25g of C is 1.25/12.01 moles = 0.1041moles. 7.00 g of Fe2O3 is 7.00/159.7 = 0.0438 moles. 2 moles of Fe2O3 react with 3 moles of C, so the ideal molar ratio of C to Fe2O3 is 3/2 = 1.5; the actual ratio is 0.1041/0.0438 = 2.375 so there is too much C and the amount of Fe2O3 will determine the amount of Fe: there will be 2 moles of Fe for each mole of Fe2O3, so there are 0.0876 mole of Fe produced, or 4.892 g

The moles of C used is 3/4 the moles of Fe produced, or (3/4)*0.0876 = 0.0657 moles C = 0.789 g of C consumed. The excess is 1.25 - 0.789 = 0.461 g C

You can also figure this by noting that the moles of C consumed is 3/2 times the moles of Fe2O3 used, or (3/2)*0.0438 = 0.0657 moles C; you started with 0.1041 moles, leaving 0.1041 - 0.0657 = 0.0384 moles C. or 0.461 g C in excess.

Answer 2

Using 1000g C, 6233.3g of Fe can be produced assuming Fe(2)o(3) is added in excess.

Using basic stoichiometry and mole ratios, no. of moles of C is 1000/12 mol.
No. of moles of Fe produced = (4/3) x (1000/12)
= 111.11 mol
Mass of Fe produced = 111.11 x (molar mass of Fe)
= 111.11 x 56.1

Theoretical yield. Fe(2)O(3) is found to be the limiting reagent. So you use the no. of moles of iron(III)oxide,

7/[(2 x 56.1) + (16 x 3)]

2 moles of iron oxide requires 3 moles of carbon.
0.43695 moles of iron oxide requires 0.065543 moles of C

Initial number of moles of carbon is 1.25/12 = 0.10416 mol
Amount of carbon left is 0.10416 - 0.065543 = 0.38616 mol.

Tell me if I was wrong. I make mistakes sometimes.

Answer 3

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Answer 4

8.064g Fe
8.75g Fe (im not sure)