Answer :
Final answer:
The reaction between nitric oxide and oxygen produces nitrogen dioxide. Using the given conditions and the ideal gas law, we find oxygen is the limiting reactant. This theoretically yields 93 g of nitrogen dioxide.
Explanation:
The stoichiometric balanced chemical equation for the reaction of nitric oxide (NO) with oxygen (O2) to form nitrogen dioxide (NO2) is 2NO(g) + O2(g) -> 2NO2(g).
Firstly, considering ideal gas law conditions, the number of moles n of a gas is given by the formula n=PV/RT, where P is pressure, V is volume, R is the universal gas constant (0.0821 L.atm/(mol.K) ) and T is temperature. In this reaction, the ratio of NO:O2:NO2 is 2:1:2.
For NO, using the given conditions (P=1.534 atm, V=35.8 L, R=0.0821 L.atm/(mol.K) and T=446K), n= (1.534 atm * 35.8 L) /(0.0821 L.atm/(mol.K) * 446K)= 1.21 moles. For O2, n= (1.534 atm * 30 L) /(0.0821 L.atm/(mol.K) * 446K)= 1.01 moles.
Next, from stoichiometric coefficients, 2 moles of NO reacts with 1 mole of O2 to yield 2 moles of NO2. Thus, O2 is the limiting reactant- we have enough NO to react fully with O2, but we'll run out of O2 first. Therefore, theoretically, 1.01 moles of O2 should produce 2*1.01= 2.02 moles of NO2.
The final step is to convert moles of NO2 into grams using the molar mass of NO2 (46.01 g/mol), giving the theoretical yield of NO2= 2.02 mol * 46.01 g/mol = 93 g of nitrogen dioxide.
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