Question

Iron

Answer 1

Answer:

110 grams

Explanation:

The given chemical equation is

$4Fe+3O_2\rightarrow 2Fe_2O_3$

$\Rightarrow 2Fe+1.5O_2\rightarrow Fe_2O_3$

2 moles of Iron, Fe, react with 1.5 moles of oxygen, $O_2$, to produce 1 mole of rust.

Or, 2x56=112 grams of Iron, Fe, reacts with 1.5x32=48 grams of oxygen, $O_2$, to produce 112+48=160 grams of rust.

[As the mass of 1 mole of Fe is 56g and the mass of 1 mole of $O_2$ is 32g]

The given mass of Iron, Fe, is 100 g.

The given mass of the oxygen, $O_2$ is 33 g.

Since 48 grams of oxygen required 112 grams of iron to react completely.

So, 1 gram of oxygen required 112/48 grams of iron to react completely.

So, 33 grams of oxygen required (112/48)x33=77 grams of iron to react completely.

Here, the availability of oxygen is less, so, the oxygen in the limiting agent.

So, the 33g of oxygen (reactant) will react with 77 g of iron (reactant) to produce 33+77=110 g of rust (product) as by the law of conservation of mass the mass of the product is the sum of masses of all the reactant. The remaining iron will remains be unreacted.