Question

19) A sample of metal ore is reacted according to the following reaction:Fe(s) +2HCL (aq) --> FeCl2(aq) + H2(g)If 24.06 mL of 5.6 M HCL are used, what mass of Fe was in the ore? Keep the answer with 2 decimal places

Answer 1

Assuming all the HCl presend in the 24.06mL reacted, we can follow the steps:

1 - Use the concentration and the volume to calculate the number of moles of HCl that reacted

2 - Apply the stoichiometry ratios to calculate the number os moles of Fe that reacted

3 - Use the tomic weight of Fe to calculate the mass of that amount of number of moles of Fe.

1 - The concentration is given by the equation:

`C=\frac{n_{\text{solute}}}{V_{\text{solution}}}`

The number of moles of solute is the same as the number of moles o HCl, because it is the solute in this case:

`\begin{gathered} C=\frac{n_(HCl)}{V_{\text{solution}}_{}} \\ n_(HCl)=C\cdot V_{\text{solution}} \end{gathered}`

So, we have:

`\begin{gathered} C=5.6mol/L \\ V_{\text{solution}}=24.06mL=24.06*10^(-3)L \\ n_(HCl)=5.6mol/L\cdot24.06*10^(-3)L=0.134736mol \end{gathered}`

2 - The coefficients of Fe and HCl are 1 and 2, respectively, so we have the following relation between their number of moles:

Fe --- HCl

1 --- 2

`\begin{gathered} (n_(Fe))/(1)=(n_(HCl))/(2) \\ n_(Fe)=(n_(HCl))/(2)=(0.134736mol)/(2)=0.067368mol \end{gathered}`

3 - The atomic weight of Fe can be checked on a periodic table:

`M_(Fe)=55.845g/mol`

So, we have:

`\begin{gathered} M_(Fe)=(m_(Fe))/(n_(Fe)) \\ m_(Fe)=n_(Fe)M_(Fe)=0.067368mol\cdot55.845g/mol=3.76216\ldots g\approx3.76g \end{gathered}`

So, there was approximately 3.76 g of Fe.