Hydrogen can react with nitrogen to form ammonia. If 1 react 11.0 grams of hydro
ID: 1054081 • Letter: H
Question
Hydrogen can react with nitrogen to form ammonia. If 1 react 11.0 grams of hydrogen with 75.0 grams of nitrogen, how many grams of NH_3 will be made? 91.2g 62.Og 153g 29.2g Hydrogen can react with nitrogen to form ammonia, if 1react 11.0 grams of hydro with 75.0 grams of nitrogen, how many grams of excess chemical are used? 11.Og 51.Og 24.0g none of the above Hydrogen can react with nitrogen to form ammonia. if 1 react 11.0 grams of by with 75.0 grams of nitrogen, how many grams of excess chemical are left over 11.0g 51.0g 24.0g none of the above What are the bond angle(s) in Xenon tetrachloride? 90, 120degree 180degree 90degree 120degree What is the hybridization of Xenon (Xe) in Xenon tetrachloride? Sp^2 sp^3 dsp^3 d^2sp^3 The BH, molecule has. Sigma bonds and piExplanation / Answer
15.
N2 + 3H2-------------> 2NH3
Amount of nitrogen = 75g
n = 75g / 28 gmol-1 = 2.67 mol
Amount of hydrogen = 11g
n = 11g / 2 gmol-1 = 5.5 mol
N2 + 3H2-------------> 2NH3
3 moles of H2 reacts with 1 mole of N2
5.5 moles of H2 reacts with 1x 5.5/3 = 1.83 moles
3 moles of H2 reacts with 1 mole of N2 to produce 2 moles of NH3
5.5 moles of H2 produce 2x5.5/3 = 3.66 moles of Ammonia
Amount of NH3 produced = 3.66 mol x 17g/mol = 62.22 g
The answer is b
16. How many grams of excess chemical are used
Limiting reagent : H2
Excess Reagent: N2
1.83 moles or N2 reacted: 1.83 mole x 28g/mol = 51.24g
The answer is b
17 .Excess reagent: 2.67 -1.83 = 0.84 mol
Amount of N2 = 0.84 mol x 28g/mol = 23.52
The answer is c
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