Answer:
N2(g) + 3H2(g) ==> 2NH3(g) ... balanced equation
Limiting reactant:
5.44 g N2 x 1 mol/28 g x 2 mol NH3/mol N2 x 17 g NH3/mol = 6.61 g NH3 formed <-- LIMITING = N2(g)
5.58 g H2 x 1 mol/2 g x 2 mol NH3/3 mol H2 x 17 g NH3/mol = 31.6 g NH3 formed
Maximum mass NH3 = 6.61 g
Excess reactant is H2(g)
Mass of excess reagent:
Initial moles H2 = 5.58 g x 1 mol/2 g = 2.79 mol
Moles H2 used up = 5.44 g N2 /28 g x 3 mol H2/mol N2 = 0.583 moles H2 used up
mass H2 remaining = 2.79 mol - 0.583 mol (x2 g/mol) = 4.41 g
I feel i did the wrong thing- I’m only in the 6th grade ;-; I’m so sry if u get this wrong I tried.
Explanation:
