Question

A student wants to reclaim the iron from an 18.0-gram sample of iron(III) oxide, which

Answer 1

Answer:

$m_{Fe}=12.6gFe$

Explanation:

Hello,

In this case, since we have grams of iron (III) oxide whose molar mass is 159.69 g/mol are able to compute the produced grams of iron by using its atomic mass that is 55.845 g/mol and their 2:4 molar ratio in the chemical reaction:

$m_{Fe}=18.0gFe_2O_3*\frac{1molFe_2O_3}{159.69gFe_2O_3}*\frac{4molFe_2O_3}{2molFe_2O_3} *\frac{55.845gFe}{1molFe_2O_3} \\\\m_{Fe}=12.6gFe$

Best regards.

Answer 2

Answer:

12.6g of Fe.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2Fe2O3(s) —> 4Fe(s) + 3O2(g)

Next, we shall determine the mass of Fe2O3 that decomposed and the mass of Fe that is produced from the balanced equation.

This is illustrated below:

Mola mass of Fe2O3 = (56 x 2) + (16x3) = 160g/mol

Mass of Fe2O3 from the balanced equation = 2 x 160= 320g

Molar mass of Fe = 56g/mol

Mass of Fe from the balanced equation = 4 x 56 = 224g

From the balanced equation above,

320g of Fe2O3 decomposes to produce 224g of Fe.

Finall, we can obtain the mass of the Fe produced from the decomposition of 18g of Fe2O3 as follow:

From the balanced equation above,

320g of Fe2O3 decomposes to produce 224g of Fe.

Therefore 18g of Fe2O3 will decompose to produce = (18x224)/320 = 12.6g of Fe.

Therefore, 12.6g of Fe is obtained from 18g of Fe2O3.