AYUDA CON ESTE EJERCICIO DE QUÍMICA (reactivo limitante), ¡POR FAVOR!
Se obtiene amoniaco mediante la reacción: N2 + H2 ---- NH3 Si se combinan 2.8 gramos de N2 y 8 gramos de H2. Calcular la masa de amoniaco obtenido.
Respuestas
Mm N2 = 28 g/mol
H2 = 2 g/mol
NH3 = 17 g/mol
1. calcular moles a partir de los gramos:
moles = masa/Mm
moles N2 = 2.8 g / 28 g/mol = 0.1 mol
mol H2 = 8 g/ 2 g/mol = 4 mol
2. calcular moles a partir de los calculos:
moles NH3 = 0.1 mol N2 x 2 mol NH3
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1 mol N2
moles NH3 = 0.2
moles NH3 = 4 mol H2 x 2 mol NH3
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3 mol H2
mol NH3 = 2.667
REACTIVO LIMITANTE: N2 (produce 0.2 mol NH3)
3. Calcular g de NH3
mol = masa/mM despejar masa
m = 0.2 mol x 17 g/mol
m = 3.4 g de NH3
N2 + 3H2 → 2NH3
Mm N2 = 28 g/mol
H2 = 2 g/mol
NH3 = 17 g/mol
1. calcular moles a partir de los gramos:
moles = masa/Mm
moles N2 = 2.8 g / 28 g/mol = 0.1 mol
mol H2 = 8 g/ 2 g/mol = 4 mol
2. calcular moles a partir de los calculos:
moles NH3 = 0.1 mol N2 x 2 mol NH3
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1 mol N2
moles NH3 = 0.2
moles NH3 = 4 mol H2 x 2 mol NH3
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3 mol H2
mol NH3 = 2.667
REACTIVO LIMITANTE: N2 (produce 0.2 mol NH3)
3. Calcular g de NH3
mol = masa/mM despejar masa
m = 0.2 mol x 17 g/mol
m = 3.4 g de NH3