Question

Iron metal reacts with chlorine gas. How many grams of FeCl3 are obtained when 515g Cl2 reacts with excess Fe.

Answer 1

Let's write down the equation of chemical reaction (unbalanced)

$$Fe + C{l_2} \to FeC{l_3}$$

and balance it (balance $Cl$ at first, then $Fe$)

$$2Fe + 3C{l_2} \to 2FeC{l_3}$$

Now we can see that for each $1$ mole of $C{l_2}$ we get $\frac{2}{3}$ moles of $FeC{l_3}$ (we divide equation by 3)

So, the number of moles of $2FeC{l_3}$ will be

$${\nu _{FeC{l_3}}} = \frac{2}{3}{\nu _{C{l_2}}}$$

Molar mass of $C{l_2}$ (atomic weight can be found in the periodic table)

$${M_{C{l_2}}} = 2 \cdot 35.45[\frac{{\text{g}}}{{{\text{mol}}}}] \approx 70.9[\frac{{\text{g}}}{{{\text{mol}}}}]$$

and ${FeC{l_3}}$

$${M_{FeC{l_3}}} = 55.85[\frac{{\text{g}}}{{{\text{mol}}}}] + 3 \cdot 35.45[\frac{{\text{g}}}{{{\text{mol}}}}] \approx 162.2[\frac{{\text{g}}}{{{\text{mol}}}}]$$

And we can use $\nu = \frac{m}{M}$ , thus

$${m_{FeC{l_3}}} = \frac{2}{3}{m_{C{l_2}}}\frac{{{M_{FeC{l_3}}}}}{{{M_{C{l_2}}}}}$$

Let's do the calculations

$${m_{FeC{l_3}}} = \frac{2}{3}515[{\text{g}}]\frac{{162.2[\frac{{\text{g}}}{{{\text{mol}}}}]}}{{70.9[\frac{{\text{g}}}{{{\text{mol}}}}]}} \approx 785.45[{\text{g}}]$$