A) 0.0194 atm
B) 0.194 atm
C) 5.82 atm
D) 0.0582 atm \[\underset{1-x}{\mathop{{{\operatorname{NH}}_{2}}COON{{H}_{4}}(s)}}\,\,\,\,\,2N{{H}_{\underset{2x}{\mathop{3}}\,}}\,(g)\,+\,\,C{{O}_{\underset{x}{\mathop{2}}\,}} (g)\] \[{{K}_{p}}={{({{\rho }_{N{{H}_{3}}}})}^{2}}\times {{\rho }_{C{{O}_{2}}}}\] \[2.9\,\,\times \,\,1{{0}^{-}}^{5}={{\left( 2x \right)}^{2}}\times x\] \[2.9\,\,\times \,\,1{{0}^{-}}^{5}=4{{x}^{3}}\] \[\operatorname{x}= 0.0194 atm\] \[\operatorname{Total} pressure = 2x +x=3x = 3\,\,\times \,\, 0.0194\] \[= 0.0582 atm\]
Correct Answer: D
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