Given the reaction: NH3(g)+O2(g)arrow NO(g)+H2O(g) What is the limiting reactant when there are .5 moles of O2 and 1.5 moles of NH3 available to react? How much excess is there? Choose matching definition Nh3, there is .5 moles of excess O2. O2, there is .moles of excess NH3. Nh3 there is 2 moles of excess Nh3. O2, there is 3 moles of excess O2.

To determine the limiting reactant, we need to compare the mole ratio of the reactants with the stoichiometric ratio in the balanced chemical equation.<br /><br />The balanced chemical equation for the reaction is:<br />$4NH3(g) + 5O2(g) \rightarrow 4NO(g) + 6H2O(g)$<br /><br />From the balanced equation, we can see that the stoichiometric ratio of $NH3$ to $O2$ is 4:5.<br /><br />Now, let's compare the given amounts of $NH3$ and $O2$ with the stoichiometric ratio:<br /><br />Given:<br />$NH3$: 1.5 moles<br />$O2$: 0.5 moles<br /><br />To find the limiting reactant, we need to calculate the amount of $O2$ required to completely react with 1.5 moles of $NH3$.<br /><br />Using the stoichiometric ratio, we can calculate the amount of $O2$ required as follows:<br />$O2$ required = $\frac{5}{4} \times NH3$ = $\frac{5}{4} \times 1.5$ = 1.875 moles<br /><br />Since we only have 0.5 moles of $O2$ available, $O2$ is the limiting reactant.<br /><br />To find the amount of excess $NH3$, we need to calculate the amount of $NH3$ that remains after the reaction is complete.<br /><br />Using the stoichiometric ratio, we can calculate the amount of $NH3$ that reacts with 0.5 moles of $O2$ as follows:<br />$NH3$ reacted = $\frac{4}{5} \times O2$ = $\frac{4}{5} \times 0.5$ = 0.4 moles<br /><br />The amount of excess $NH3$ is then:<br />$NH3$ excess = $NH3$ initial - $NH3$ reacted = 1.5 - 0.4 = 1.1 moles<br /><br />Therefore, the correct answer is:<br />$NH3$, there is 1.1 moles of excess $NH3$.