Answer:
[tex]m_{CO2}=8.33gCO2[/tex]
Explanation:
Hello.
In this case, since the chemical reaction is:
[tex]C_7H_5N_3O_6(s)+\frac{21}{4} O2(g)\rightarrow 7CO_2(g)+\frac{3}{2} N_2(g)+\frac{5}{2} H_2O(l)[/tex]
Thus, since we have the initial moles of TNT:
[tex]n_{TNT}=25.7g*\frac{1mol}{227.13g} =0.113mol[/tex]
And the initial moles of oxygen given the ideal gas equation:
[tex]n_{O_2}=\frac{6.93atm*0.5L}{0.082\frac{atm*L}{mol*K}*298K}=0.142molO_2[/tex]
Given the 1:7 mole ratio between TNT and carbon dioxide and the 21/4:7 mole ratio between oxygen and carbon dioxide, we compute the yielded moles by each reactant:
[tex]n_{CO_2}^{by\ TNT}=0.113molTNT*\frac{7molCO_2}{1molTNT}=0.791molCO_2\\\\ n_{CO_2}^{by\ O_2}=0.142molO_2*\frac{7molCO_2}{21/4molO_2}=0.189molCO_2[/tex]
Thus, since oxygen yields less moles of carbon dioxide than TNT, we infer it is the limiting reactant, therefore, the produced mass of carbon dioxide is:
[tex]m_{CO2}=0.189molCO2*\frac{44.01gCO2}{1molCO2}\\\\m_{CO2}=8.33gCO2[/tex]
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