## Related questions with answers

Dinitrogen pentoxide decomposes into nitrogen dioxide and oxygen. If 5.00 L of

$\mathrm { N } _ { 2 } \mathrm { O } _ { 5 }$

reacts at STP, what volume of

$\mathrm { NO } _ { 2 }$

is produced when measured at

$64.5 ^ { \circ } C$

and 1.76 atm?

Solution

VerifiedThe coefficients in a chemical reaction involving gases indicate the relative numbers of molecules, the relative numbers of moles, and the relative volumes. All volumes are to be compared at the same temperature and pressure. Therefore, volume ratios can be used like mole ratios to find the unknowns.

$\mathrm{N_2O_5 \rightarrow NO_2 + \dfrac{1}{2} \text{ }O_2}$

$V_{NO_2} = \mathrm{ 5 \text{ }L \text{ }N_2O_5 \text{ x } (\dfrac{2 \text{ }NO_2 }{ N_2O_5 })}$

$\bold{V_{NO_2} = 10 \text{ }L \text{ }NO_2}$

The combined gas law expresses the relationship between pressure, volume, and temperature of a fixed amount of gas.

$\dfrac{PV}{T}=k$

In the equation, k is constant and depends on the amount of gas. The combined gas law can also be written as follows:

$\dfrac{ P_1V_1}{T_1}= \dfrac{ P_2V_2}{T_2}$

For purposes of comparison, scientists have agreed on standard conditions of 1 atm pressure and $0^oC$.

$V_2= \dfrac{P_1V_1T_2}{P_2T_1}$

$V_2= \mathrm{\dfrac{(1 \text{ }atm) \text{ x } (10 \text{ }L) \text{ x } (64.5+273.15) \text{ }K}{(1.76 \text{ }atm) \text{ x } (0+273.15) \text{ }K}}$\

$\bold{V_2= 7.0131 \text{ } L \text{ } NO_2}$

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