Respuesta :
Amount of nitrogen produced = 1.624 g
To calculate the moles, we use the equation:
4NH3(g) + 3 O2 (g) ---> 2 N2 (g) + 6 H2O (g) ....(1)
For ammonia:
Putting values in above equation, we get:
Moles of ammonia = 7.00 x 1022 / 6.023 * 10 ^23 = 0.116 mol
Moles of oxygen = 6.00 x 1022 / 6.023 * 10 ^23 = 0.099 mol
For the reaction:
4NH3(g) + 3 O2 (g) ---> 2 N2 (g) + 6 H2O (g)
By Stoichiometry of the reaction,
4 moles of ammonia combine with 3 moles of Oxygen
Thus 0.116 moles of ammonia will combine with= 3/4 * 0.116 = 0.087 of oxygen
Thus ammonia is the limiting reagent as it limits the formation of product.
4 moles of ammonia produces 2 moles of nitrogen
0.132 moles of ammonia will produce=2/4 * 0.116 = 0.058 of nitrogen
Molar mass of nitrogen = 28 g/mol
Amount of nitrogen produced=
No of moles * molar mass = 0.058 * 282 = 1.624 g
To learn more about molecular mass from the given link
https://brainly.com/question/15428819
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