Question
Consider the formation of nitrogen dioxide from nitric oxide and oxygen:$$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)$$If $9.0 \mathrm{L}$ of $\mathrm{NO}$ is combined with excess $\mathrm{O}_{2}$ at $\mathrm{STP}$, what is the volume in liters of the $\mathrm{NO}_{2}$ produced?
Step 1
We know that at standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 L. So, the number of moles of NO can be calculated as follows: \[n_{NO} = \frac{V_{NO}}{V_{STP}} = \frac{9.0 \, L}{22.4 \, L/mol} = 0.401 \, mol\] Show more…
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Consider the formation of nitrogen dioxide from nitric oxide and oxygen: $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)$$ If $9.0 \mathrm{L}$ of $\mathrm{NO}$ are reacted with excess $\mathrm{O}_{2}$ at $\mathrm{STP}$ what is the volume in liters of the $\mathrm{NO}_{2}$ produced?
Consider the formation of nitrogen dioxide from nitric oxide and oxygen: 2NO(g) + O2(g) --> 2NO2(g). If 9.0 L of NO are reacted with excess O2 at STP, what is the volume in liters of the NO2 gas produced? (Recall that the gas constant, R, = 0.082057 L-atm/mol-K). Please show work.
Nitrogen monoxide reacts with oxygen to give nitrogen dioxide. $$2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})$$ (a) If you mix $\mathrm{NO}$ and $\mathrm{O}_{2}$ in the correct stoichiometric ratio, and NO has a volume of $150 \mathrm{mL}$, what volume of O, is required (at the same pressure and temperature)? (b) After reaction is complete between 150 mL of $\mathrm{NO}$ and the stoichiometric volume of $\mathrm{O}_{2},$ what is the volume of $\mathrm{NO}_{2}$ (at the same pressure and temperature)?
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