454546 points Save Answer Under certain conditions the react

4.54546 points Save Answer Under certain conditions the reaction of ammonia with excess oxygen will produce a 295% yield of NO. What mass of NH3 must react with excess oxygen to yield 157 g NO? 4NHs(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) 0 a. 89.1 g O b.302 g O c. 263 g Od. 26.3 g O e.938 g

Solution

The balanced reaction will be

4NH3(g) + 5O2(g) ------- 4NO(g) + 6H2O(g)

Molar mass of NO = 14 + 16 = 30 gm/mol

Number of moles of NO = Mass/molar mass = 157/30 = 5.23333 moles

4 moles of NH3 gives 4 moles of NO in the balanced reaction, hence number of moles of NH3 required if 100% yield = 5.23333 moles

Number of moles of NH3 * Percentage Yield = Number of moles of NO

Number of moles of NH3 = 100 * 5.233333/29.5 = 17.74 moles

Molar mass of NH3 = 14 + 3 * 1 = 17 gm/mol

Mass of NH3 = 17.74 mol * 17 gm/mol = 301.58 grams

Hence the correct answer must be Option B (302 grams)