Consider the following balanced chemical equation: 2Fe(s)+3Pb(NO_(3))_(2)(aq)arrow 3Pb(s)+2Fe(NO_(3)) If 3.6 moles of Fe are reacted, how many moles of Pb(NO_(3))_(2) were reacted? square

To determine how many moles of \( Pb(NO_3)_2 \) were reacted when 3.6 moles of Fe are reacted, we need to use the stoichiometric relationship from the balanced chemical equation:<br /><br />\[ 2Fe(s) + 3Pb(NO_3)_2(aq) \rightarrow 3Pb(s) + 2Fe(NO_3)_3 \]<br /><br />From the equation, we see that 2 moles of Fe react with 3 moles of \( Pb(NO_3)_2 \).<br /><br />We can set up a proportion to find the number of moles of \( Pb(NO_3)_2 \) that react with 3.6 moles of Fe:<br /><br />\[<br />\frac{2 \text{ moles of Fe}}{3 \text{ moles of } Pb(NO_3)_2} = \frac{3.6 \text{ moles of Fe}}{x \text{ moles of } Pb(NO_3)_2}<br />\]<br /><br />Now, solve for \( x \):<br /><br />\[<br />x = \frac{3.6 \text{ moles of Fe} \times 3 \text{ moles of } Pb(NO_3)_2}{2 \text{ moles of Fe}}<br />\]<br /><br />\[<br />x = \frac{10.8 \text{ moles of } Pb(NO_3)_2}{2}<br />\]<br /><br />\[<br />x = 5.4 \text{ moles of } Pb(NO_3)_2<br />\]<br /><br />Therefore, 5.4 moles of \( Pb(NO_3)_2 \) were reacted.