What is the % yield when 4.00 g of H 2 (g)reacts with 45.0 g of N 2 (g) and form

Question

What is the % yield when 4.00 g of H2(g)reacts with 45.0 g of N2(g) and forms 20.0 g ofNH3(g)? What is the % yield when 4.00 g of H2(g)reacts with 45.0 g of N2(g) and forms 20.0 g ofNH3(g)?

Explanation / Answer

Equation: 3H2 + N2 -> 2NH3 Find the limiting reagent: 4.00 g H2 x (1 mol / 2 g) x (2 mol NH3 / 3 mol H2) x (17 g / 1 molNH3) = 22.67 grams NH3 45.0 g N2 x (1 mol/ 28 g) x (2 mol NH3 / 1 mol N2) x (17 g / 1molNH3) = 54.64 grams NH3 Since H2 yields less NH3: H2 is your limiting reagent, and theamoung is 22.67 grams NH3 Percent yield = actual/predicted x 100 Percent yield = 20.0 g / 22.67 g x 100 = 88.2%

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