4NH3+5O2->4NO+6H2O a)what are the oxidation numbers for NH3 and NO? What is the

Question

4NH3+5O2->4NO+6H2O a)what are the oxidation numbers for NH3 and NO? What is the reducing agent in this reaction b) when 8 moles of NH3 are combined with 10 moles of O2 in a rigid container the initial pressure before a reaction occurs is 3.2 atm. What will be the pressure inside the container after the reaction occurs if the temperature remains constant (above the boiling temperature of water)? Assume 100% yield. c) The container from part b is cooled 5 C causing all the water vapor to condense to a liquid- the other gases remain gases. The mass of the liquid is 212 g. What is the true % yield of the reaction? d) The total pressure of the container after the water condenses (5C) is .857 atm. What are the partial pressures of the 4 gases at 5C. e)What was the temperature of the container in Kelvin before it was cooled to 5C?

Explanation / Answer

(a) Usually, the oxidation number of H and O are +1 and -2, respectively, in chemical compounds. Since NH3 and NO are neutral, N in NH3 is in the -3 oxidation state and N in NO is in the +2 oxidation state. We see that O goes from 0 oxidation state in O2 to -2 oxdiation state in NO. Therefore, NH3 is the reducing agent (it "reduces" the oxidation state/number of O in O2 during the reaction). (b) Assuming 100% yield, we see that the reaction stoichiometry indicates that NH3 reacts with O2 in 4:5 reactions, consistent with the condition given (8:10 -> 4:5). No there is no limiting reagent here. Following stoichiometry, we see that 8 moles of NO is produced, and 12 moles of H2O is produced. Apply Ideal Gas Law: PV = nRT, where V, R, T are equal, which left us P = n*(RT/V) -> P1 = n1 * (RT/V) & P2 = n2 * (RT/V). Therefore P1/P2 = n1/n2 = (8+10) / (8+12) = 9/10 ; P2 = P1 *n2/n1 = 3.2 * 10/9 = 3.5556 atm. (c) The molar mass of water is 18g/mole. In 212 g of water, we have 11.77778 moles of water produced from the reaction. Comparing to case (b) where 12 moles of water is produced in 100% condition, we see that the actual yield is 98.148% (d) There are four gases present, unreacted NH3 and O2 gases, as well as reaction product NO and H2O gases. Even though water vapor is condensed to liquid form at 5C, there might still be saturated vapor in the gas phase. However, absolute saturated vapor pressure at 5C is almost negligible (~0). From (c), we see that the actual yield is about 98.148%, this means that around 7.8518 moles of NO is produced, and around 0.14816 moles of NH3 is left ( which is 100-98.148= 1.852% of the original amount), there is also 0.1852 moles of unreacted O2 left. The ratio between the four gases is then 7.8518: 0.14816: 0.1852. "Dividing" the total pressure among the species according to their mole fraction (ideal gas law is used again here). The partial pressures are 0.822 atm, 0.0155 atm, and 0.01939 atm for NO, NH3 and O2 respectively. (e) Apply ideal gas law once again. PV = nRT -> T = PV/nR, where V and R are constant. Therefore. T1 = (P1/n1) * (V/R) and T2 = (P2/n2) * (V/R) -> T1 = T2 * (P1/P2) * (n2/n1 ) = (3.5556 / 0.857)*(8.18516/19.96294) = (5 + 273.15) K * 1.7 = 473.166 K

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