91Ó°ÊÓ

We are given a range of different scenarios with different amounts of molecules and moles of the reactants. For each scenario, we will compare the given ratio of \(\mathrm{H}_2\) and \(\mathrm{O}_2\) to the stoichiometric ratio in the balanced equation.

For each scenario, we'll calculate the ratios of the given reactants to the needed stoichiometric ratio and identify the limiting reagent. a. For 50 molecules of \( \mathrm{H}_{2} \) and 25 molecules of \( \mathrm{O}_{2} \): We can see that there are enough molecules of \( \mathrm{O}_{2} \) present for the reaction to proceed entirely. Therefore, the limiting reagent in this case is \( \mathrm{H}_{2} \). b. For 100 molecules of \( \mathrm{H}_{2} \) and 40 molecules of \( \mathrm{O}_{2} \): Divide the amount of molecules by the stoichiometric coefficients (2 for \( \mathrm{H}_{2} \) and 1 for \( \mathrm{O}_{2} \)): \[\frac{100}{2} : \frac{40}{1} = 50:40\] Since there are more available moles of \( \mathrm{H}_{2} \) compared to the stoichiometric ratio, the limiting reagent is \( \mathrm{O}_{2} \). c. For 100 molecules of \( \mathrm{H}_{2} \) and 100 molecules of \( \mathrm{O}_{2} \): Divide the molecules by the stoichiometric coefficients to get a ratio of 50:100. Thus, the limiting reagent in this case is \( \mathrm{H}_{2} \). d. For \( 0.50 \ \mathrm{mol} \ \mathrm{H}_{2} \) and \( 0.75 \ \mathrm{mol} \ \mathrm{O}_{2} \): Calculate the ratio of the moles: \[\frac{0.50}{2}: \frac{0.75}{1} = 0.25:0.75\] Since there are fewer available moles of \(\mathrm{H}_2\) as compared to the stoichiometric ratio, the limiting reagent is \( \mathrm{H}_{2} \). e. For \( 0.80 \ \mathrm{mol} \ \mathrm{H}_{2} \) and \( 0.75 \ \mathrm{mol} \ \mathrm{O}_{2} \): Calculate the ratio of the moles: \[\frac{0.80}{2}: \frac{0.75}{1} = 0.40:0.75\] Since there are fewer available moles of \(\mathrm{H}_2\) as compared to the stoichiometric ratio, the limiting reagent is \( \mathrm{H}_{2} \). f. For \((1.0 \ \mathrm{g} \ \mathrm{H}_{2})\) and \((0.25 \ \mathrm{mol} \ \mathrm{O}_{2})\): First, convert the grams of \( \mathrm{H}_{2} \) to moles (molar mass of \( \mathrm{H}_{2} = 2 \ \mathrm{g/mol} \)): \[\frac{1.0 \mathrm{~g} \mathrm{H}_{2}}{2 \mathrm{~g} \mathrm{H}_{2 / mol}} = 0.50 \ \mathrm{mol} \ \mathrm{H}_{2} \] Now calculate the ratio of the moles: \[\frac{0.50}{2}: \frac{0.25}{1} = 0.25:0.25\] The available amounts of \(\mathrm{H}_2\) and \(\mathrm{O}_2\) are equal to the stoichiometric ratio, meaning that in this case there is no limiting reagent. g. For \( 5.00 \ \mathrm{g} \ \mathrm{H}_{2} \) and \( 56.00 \ \mathrm{g} \ \mathrm{O}_{2} \): First, convert the grams of \( \mathrm{H}_{2} \) and \( \mathrm{O}_{2} \) to moles (molar mass of \( \mathrm{O}_{2} = 32 \ \mathrm{g/mol} \)): \[ \frac{5.00 \ \mathrm{g} \ \mathrm{H}_{2}}{2 \ \mathrm{g/mol}} = 2.5 \ \mathrm{mol} \ \mathrm{H}_{2}\] \[ \frac{56.00 \ \mathrm{g} \ \mathrm{O}_{2}}{32 \ \mathrm{g/mol}} = 1.75 \ \mathrm{mol} \ \mathrm{O}_{2} \] Calculate the ratio of the moles: \[\frac{2.5}{2}: \frac{1.75}{1} = 1.25:1.75\] Since there are fewer available moles of \(\mathrm{H}_2\) as compared to the stoichiometric ratio, the limiting reagent is \( \mathrm{H}_{2} \).