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############################# Video Source: www.youtube.com/watch?v=W9jC5mFQAt4

En este vídeo de estequiometría se explica como se obtienen los gramos de un producto a partir de los gramos de un reactivo, conociendo sus pesos moleculares y la reacción Química. • Tutoriales de química, explicaciones, conceptos, ejercicios, simuladores,ejemplos, aplicaciones • A los químicos les interesa conocer la masa de reactivos que necesitan para obtener una cantidad de producto determinada en una reacción química, o la cantidad de producto que pueden obtener a partir de una determinada cantidad de reactivos. Los cálculos que hay que hacer para resolver estas cuestiones se llaman cálculos estequiométricos. • Para realizar los cálculos estequiométricos es necesario disponer de la ecuación química ajustada de la reacción. Entonces podemos conocer la cantidad de moléculas de un producto que se puede obtener a partir de una cierta cantidad de moléculas de los reactivos. Por ejemplo con 2 moléculas de hidrógeno (H2) y 1 molécula de oxígeno (O2) se pueden obtener 2 moléculas de agua (H2O). Si sabemos la masa de cada molécula sabemos también la relación entre las masas de reactivos y productos en la reacción. Estas masas si que las conocemos. Se llaman masas moleculares, y se calculan sumando las masas de los átomos que componen las moléculas, las masas atómicas. Estas las encontrarás en cualquier tabla periódica expresada en u (unidades de masa atómica). Pero como puedes imaginar son masas muy pequeñas, del orden de los 10-24 g. Por eso los químicos han definido una nueva unidad para medir el número de partículas (átomos o moléculas), a la que han llamado mol y que se define así: • n mol de una sustancia es una cantidad equivalente a la que representa su masa atómica en umas expresada en gramos. En un mol de una sustancia hay 6,022 . 10 exponente 23 partículas (átomos, moléculas, iones...) • Así, la relación en moles de moléculas en nuestra reacción entre el hidrógeno y el oxígeno también viene dada por los coeficientes estequiométricos, de manera que también la podemos leer como: • 2H2 + O2------------------------} 2H2O • 2 moles de moléculas de hidrógeno reaccionan con 1 mol de moléculas de oxígeno para dar 2 moles de moléculas de agua • o, sabiendo que las masas atómicas del hidrógeno y del oxígeno son: • M (H) = 1 u M(O) = 16 u • y que por lo tanto las masas moleculares del gas hidrógeno, del gas oxígeno y del agua son: • M (H2) = 2 . M (H) = 2 . 1 u = 2 u • M (O2) = 2 . M (O) = 2 . 16 u = 32 u • M (H2O) = 2 . M (H) + 1 . M (O) = 2 . 2 u + 1 . 16 u = 18 u • de manera que la masa de 1 mol de cada sustancia será: • M (H2) = 2 g/mol • M (O2) = 32 g/mol • M (H2O) = 18 g/mol • • podemos leer la ecuación química ajustada de la reacción como: • 4 g de hidrógeno reaccionan con 32 g de oxígeno para dar 36 g de agua . Observa que la suma de las masas de los reactivos es igual a la suma de las masas de los productos, como tenía que ser (ley de 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