N2(g) + 3H2(g) ? 2NH3(g) 1. What is the maximum mass of ammonia that can be prod

Question

N2(g) + 3H2(g) ? 2NH3(g) 1. What is the maximum mass of ammonia that can be produced from a mixture of 194.3 g of N2 and 45.29 g of H2? 2. Which element would be left partially unreacted? (either nitrogen or hydrogen) 3. What mass of the starting material would remain unreacted? My Approach to the problem: 194.3 g N2/ 28.0134 g/mol = 6.936 mol of N2 45.29 g H2/ 2.0159 = 22.47 mol of H2 -- I determined Nitrogen is the limiting reactant -- 2 * 6.936 = 13.87 mol of NH3 * 17.0306 g/mol = 236.2 g of NH3 produced I then proceded to calculate the mass of the ureacted material by doing: (22.47 mol of H2 - (6.936 mol of N2 * 3/1)) 2.016 = 3.35 g However my online assignment is marking it wrong. Did I do any mistakes in my calculations?

Explanation / Answer

N2 (Molecular weight) = 28.0134 g/mole
H2(Molecular weight) = 2.0158 g/mole
NH3 (Molecular weight) = 17.0304 g/mole

N2(g) + 3H2(g) --> 2NH3(g)

194.3 g N2 = 194.3 g/28.0134g

= 6.935 mole N2

45.29 g H2 = 45.29 g /2.0158g

= 22.46 mole H2

6.935 moles N2 need 45.29*6.935 = 314 mole H2

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