8.5 g sample of NH3 on oxidation produces 4.5 g of NO. Calculate the percent yield?

Question

Reaction: 4 NH3 + 5 O2 > 4 NO + 6 H2O

Answer 1

Molar mass of NH3 = 17
Molar mass of NO = 30

Moles of NH3 = 0.5
Moles of NO = 0.15

Thus %age yield = 30%

Answer 2

2H20 + 2Na >> 2NaOH + H2 sodium is in extra, so water is the proscribing reagent and also you want to remodel the eighty.0 g of H20 into moles utilizing the molar mass of water. eighty.0 g H20 x (a million mole H20/18.01 g H20)= 4.4 mol H20 then you truly use the mole to mole ratio of the reaction. 2 moles of water are equivalent to at least a million mole of H2: 4.4 mol H20 x( a million mol H2/ 2 mol H20)=2.2 mol H2. then convert H2 back to grams which outcomes being 4.40 4 and use the % yield equation: (truly yield/theoretical yield) X one hundred. answer: (4.14 g H2/ 4.40 4 g H2) x one hundred= ninety 3.2% yield