Dada la siguiente reacción:
\qquad
R1_RATIO === 1 ? "" : R1_RATIOR1 +
R2_RATIO === 1 ? "" : R2_RATIOR2 \rightarrow
P1_RATIO === 1 ? "" : P1_RATIOP1
+ P2_RATIO === 1 ? "" : P2_RATIOP2
¿Cuántos gramos de P1 se producirán con R1_MASS \text{g} de R1 y R2_MASS \text{g} de R2?
0.75 cualquier respuesta dentro de 1 gram será aceptada para permitter el redondeo
\dfrac{R1_MASS \cancel{\text{g}}}{R1_MOLAR_MASS \cancel{\text{g}} / \text{mol}} =
\blue{\text{ R1_MOL plural_form(MOLE, R1_MOL)}} \text{ OF }R1
[Explicar]
Primero queremos convertir la cantidad dada de R1 de gramos a moles. Para hacer esto, dividimos la cantidad dada de R1 entre la masa molecular del R1.
\dfrac{\text{GRAMS_OF }R1}{\text{MOLAR_MASS_OF }R1} = \text{MOLES_OF }R1
Para encontrar la masa molecular de R1, buscamos el peso atómico de cada átomo en la molécula de R1 en la tabla periódica y los sumamos. En este caso, es R1_MOLAR_MASS \text{g/mol}.
Dividing the given R1_MASS \text{g} of R1 by the molar mass of
R1_MOLAR_MASS \text{g/mol} tells us we're starting with
R1_MOL\text{ plural_form(MOLE, R1_MOL) OF }R1.
\dfrac{R2_MASS \cancel{\text{g}}}{R2_MOLAR_MASS \cancel{\text{g}} / \text{mol}} =
\green{\text{ plural(R2_MOL, "mole")}} \text{ OF }R2
[Explicar]
Queremos convertir la cantidad dada de R2 de gramos a moles. Para hacer esto, dividimos la cantidad dada de R2 entre la masa molecular del R2.
\dfrac{\text{GRAMS_OF }R2}{\text{MOLAR_MASS_OF }R2} = \text{MOLES_OF }R2
Para encontrar la masa molecular de R2, buscamos el peso atómico de cada átomo en la molécula de R2 en la tabla periódica y los sumamos. En este caso, es R2_MOLAR_MASS \text{g/mol}.
Dividing the given R2_MASS \text{g} of R2 by the molar mass of
R2_MOLAR_MASS \text{g/mol} tells us we're starting with
\text{ R2_MOL plural_form(MOLE, R2_MOL)} \text{ OF }R2.
La razón de moles de \dfrac{R1}{R2} en la reacción es \dfrac{R1_RATIO}{R2_RATIO}. [Explicar]
The reaction is \blue{R1_RATIO}R1 +
\red{R2_RATIO}R2 \rightarrow
P1_RATIOP1 + P2_RATIOP2.
The coefficients in front of each molecule tell us in what ratios the molecules react. In this case
cardinalThrough20(R1_RATIO) R1 for every
cardinalThrough20(R2_RATIO) R2 molecule
\qquad
\dfrac{R1}{R2} = \dfrac{R1_RATIO}{R2_RATIO} =
\dfrac{\blue{\text{ R1_MOL plural_form(MOLE, R1_MOL)}}}{x}
[Mostrar enfoque alernativo]
\dfrac{R1}{R2} = \dfrac{R1_RATIO}{R2_RATIO} =
\dfrac{x}{\green{\text{ plural(R2_MOL, "mole")}}}
Instead of finding out how much R2 we need to react with all of our R1, we could
figure out how much R1 we need to react with all of our R2. In this case,
x = \text{ roundTo(3, R2_MOL * R1_RATIO / R2_RATIO) plural_form(MOLE, roundTo(3, R2_MOL * R1_RATIO / R2_RATIO))} of
R1 needed, which
is more than we have. Therefore R1 is the limiting reagent.
x = \text{ roundTo(3, R1_MOL * R2_RATIO / R1_RATIO) plural_form(MOLE, roundTo(3, R1_MOL * R2_RATIO / R1_RATIO))} of
R2 needed.
We have \text{ R2_MOL plural_form(MOLE, R2_MOL)} of R2, which is more
than we need. Therefore R1 is the limiting reagent.
La razón de moles de \dfrac{R1}{P1} en la reacción es \dfrac{R1_RATIO}{P1_RATIO}. [Explicar]
The reaction is \blue{R1_RATIO}R1 +
R2_RATIOR2 \rightarrow
\red{P1_RATIO}P1
+ P2_RATIOP2.
The coefficients in front of each molecule tell us in what ratios the molecules react. In this case
cardinalThrough20(R1_RATIO) R1 for every
cardinalThrough20(P1_RATIO) P1 molecule.
\qquad
\dfrac{R1}{P1} = \dfrac{R1_RATIO}{P1_RATIO} =
\dfrac{\blue{\text{ R1_MOL plural_form(MOLE, R1_MOL)}}}{x}
x = \text{ P1_MOL plural_form(MOLE, P1_MOL)} of P1 produced.
La razón de moles de \dfrac{R1}{R2} en la reacción es \dfrac{R1_RATIO}{R2_RATIO}. [Explicar]
The reaction is \blue{R1_RATIO}R1 +
\red{R2_RATIO}R2 \rightarrow
P1_RATIOP1 + P2_RATIOP2.
The coefficients in front of each molecule tell us in what ratios the molecules react. In this case
cardinalThrough20(R1_RATIO) R1 for every
cardinalThrough20(R2_RATIO) R2 molecule.
\qquad
\dfrac{R1}{R2} = \dfrac{R1_RATIO}{R2_RATIO} =
\dfrac{x}{\green{\text{ R2_MOL plural_form(MOLE, R2_MOL)}}}
\qquad
[Mostrar enfoque alernativ]
\dfrac{R1}{R2} = \dfrac{R1_RATIO}{R2_RATIO} =
\dfrac{\blue{\text{ R1_MOL plural_form(MOLE, R1_MOL)}}}{x}
Instead of finding out how much R1 we need to react with all of our R2, we could
figure out how much R2 we need to react with all of our R1. In this case,
x = \text{ roundTo(3, R1_MOL * R2_RATIO / R1_RATIO) plural_form(MOLE, roundTo(3, R1_MOL * R2_RATIO / R1_RATIO))} of
R2 needed, which
is more than we have. Therefore R2 is the limiting reagent.
x = \text{ roundTo(3, R2_MOL * R1_RATIO / R2_RATIO) plural_form(MOLE, roundTo(3, R2_MOL * R1_RATIO / R2_RATIO))} of
R1 needed.
We have \text{ R1_MOL plural_form(MOLE, R1_MOL)} of R1, which is more
than we need. Therefore R2 is the limiting reagent.
La razón de moles de \dfrac{R2}{P1} en la reacción es \dfrac{R2_RATIO}{P1_RATIO}. [Explicar]
The reaction is R1_RATIOR1 +
\blue{R2_RATIO}R2 \rightarrow
\red{P1_RATIO}P1
+ P2_RATIOP2.
The coefficients in front of each molecule tell us in what ratios the molecules react. In this case
cardinalThrough20(R2_RATIO) R2 for every
cardinalThrough20(P1_RATIO) P1 molecule.
\qquad
\dfrac{R2}{P1} = \dfrac{R2_RATIO}{P1_RATIO} =
\dfrac{\green{\text{ R2_MOL plural_form(MOLE, R2_MOL)}}}{x}
x = \text{ P1_MOL plural_form(MOLE, P1_MOL)} of P1 produced.
\cancel{\text{P1_MOL plural_form(MOLE, P1_MOL)}}
P1 \times \dfrac{P1_MOLAR_MASS \text{g}}{\cancel{\text{plural_form(MOLE, 1)}}} =
\text{ P1_MASS plural_form(GRAM, P1_MASS)} \text{ OF }P1