Ammonia reacts with oxygen to form nitric oxide and water vapor: 4NH3 + 5O2 ? 4NO + 6H2O When 40.0 g NH3 and 50.0 g O2 are allowed to react, which is the limiting reagent? NH3 O2 NO H2O No reagent is limiting.
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The molar mass of NH3 is approximately 17.03 g/mol, and the molar mass of O2 is approximately 32.00 g/mol. So, the number of moles of NH3 is 40.0 g / 17.03 g/mol = 2.35 mol, and the number of moles of O2 is 50.0 g / 32.00 g/mol = 1.56 mol. According to the Show more…
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Ammonia reacts with oxygen to form nitric oxide and water vapor: 4NH3 + 5O2 → 4NO + 6H2O When 20.0 g NH3 and 50.0 g O2 are allowed to react, which is the limiting reagent? How much NO will be produced as a result of the reaction? * Must show work.
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Ammonia reacts with oxygen to form nitric oxide and water vapor: 4NH3 + 5O2 ? 4NO + 6H2O When 20.0 g NH3 and 50.0 g O2 are allowed to react, which is the limiting reagent (reactant)? How many grams of excess reactant are left over after the reaction is completed?
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One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of $\mathrm{NH}_{3}$ to $\mathrm{NO} :$ $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ In a certain experiment, 2.00 $\mathrm{g}$ of $\mathrm{NH}_{3}$ reacts with 2.50 $\mathrm{g}$ of $\mathrm{O}_{2} .$ (a) Which is the limiting reactant? (b) How many grams of $\mathrm{NO}$ and $\mathrm{H}_{2} \mathrm{O}$ form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.
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