Question

a 3.56 g sample of iron powder was heated in gaseous chlorine,and 10.39 g of an iron chloride was formed. what is the percent composition of the compound? show the reaction equation.

Answer 1

A 3.56 g sample of iron powder was heated in gaseous chlorine, and 10.39 g of an iron chloride was formed. what is the percent composition of the compound? show the reaction equation.

Solution:

$$2 \mathrm{Fe} + 3 \mathrm{Cl_2} = 2 \mathrm{FeCl_3}$$

According to the reaction equation is a lot of chlorine:

$$\frac{m (\mathrm{Cl_2})}{3 * M (\mathrm{Cl_2})} = \frac{m (\mathrm{Fe})}{2 * M (\mathrm{Fe})}$$ $$m (\mathrm{Cl_2}) = \frac{3 * M (\mathrm{Cl_2}) * m (\mathrm{Fe})}{2 * M (\mathrm{Fe})}$$ $$m (\mathrm{Cl_2}) = \frac{3 * 71 * 3,56}{2 * 55,8} = 6.79 \, \mathrm{g}$$

Find the amount of the substance chlorine and iron, have reacted:

$$n (\mathrm{Cl_2}) = \frac{m (\mathrm{Cl_2})}{M (\mathrm{Cl_2})} = \frac{6,79}{71} = 0,1 \, \mathrm{mol}$$ $$n (\mathrm{Fe}) = \frac{m (\mathrm{Fe})}{M (\mathrm{Fe})} = \frac{3,56}{55,8} = 0,06 \, \mathrm{mol}$$

Based on the number of iron-reacted, we received a lot of practice of iron chloride:

$$n (\mathrm{Fe}) = n (\mathrm{FeCl_3})$$ $$m (\mathrm{FeCl_3}) = n (\mathrm{FeCl_3}) * M (\mathrm{FeCl_3})$$ $$m (\mathrm{FeCl_3}) = 0,06 * 162,5 = 9,75 \, \mathrm{g}$$ $$w (\mathrm{FeCl_3}) = \frac{m (\mathrm{FeCl_3})}{m' (\mathrm{FeCl_3})}$$ $$w (\mathrm{FeCl_3}) = \frac{9,75}{10,39} = 0,94 = 94\%$$

Answer: w (FeCl₃) = 0,94 = 94%