Consider the following balanced chemical equation: 2Fe(s)+6HCl(aq)arrow 3H_(2)(g)+2FeCl_(3)(aq) If 1.8 moles of FeCl_(3) are formed, how many moles of H_(2) were formed at standard temperature and pressure? square

To determine the number of moles of \( H_2 \) formed when 1.8 moles of \( FeCl_3 \) are formed, we need to use the stoichiometric relationship from the balanced chemical equation:<br /><br />\[ 2Fe(s) + 6HCl(aq) \rightarrow 3H_2(g) + 2FeCl_3(aq) \]<br /><br />From the equation, we see that 2 moles of \( FeCl_3 \) are produced for every 3 moles of \( H_2 \) produced. This gives us the following ratio:<br /><br />\[ \frac{3 \text{ moles } H_2}{2 \text{ moles } FeCl_3} \]<br /><br />We can set up a proportion to find the number of moles of \( H_2 \) produced when 1.8 moles of \( FeCl_3 \) are formed:<br /><br />\[ \frac{3 \text{ moles } H_2}{2 \text{ moles } FeCl_3} = \frac{x \text{ moles } H_2}{1.8 \text{ moles } FeCl_3} \]<br /><br />Now, solve for \( x \):<br /><br />\[ x = \frac{3 \text{ moles } H_2}{2 \text{ moles } FeCl_3} \times 1.8 \text{ moles } FeCl_3 \]<br /><br />\[ x = \frac{3 \times 1.8}{2} \]<br /><br />\[ x = \frac{5.4}{2} \]<br /><br />\[ x = 2.7 \text{ moles } H_2 \]<br /><br />Therefore, 2.7 moles of \( H_2 \) were formed at standard temperature and pressure when 1.8 moles of \( FeCl_3 \) are formed.