Answer:
88.98 %.
Explanation:
- From the balanced equation:
<em>2HCl + Pb(NO₃)₂ → 2HNO₃ + PbCl₂</em>
- It is clear that 1.0 mole of Pb(NO₃)₂ reacts with 2.0 moles of HCl to produce 1.0 mole of PbCl₂ and 2.0 moles of HNO₃.
- <em>The percent yield % of lead(II) chloride (PbCl₂) = [(actual yield) / (calculated yield)] x 100.</em>
- The actual yield of lead(II) chloride (PbCl₂) = 650 g.
- Now, we need to calculate the calculated yield of lead(II) chloride (PbCl₂).
- We need to calculate the no. of moles (n) of lead(II) nitrate (Pb(NO₃)₂) (870 grams) using the relation: <em>n = mass / molar mass.</em>
- n of lead(II) nitrate (Pb(NO₃)₂) = mass / molar mass = (870 g) / (331.2 g/mol) = 2.63 mol.
- Since HCl is in excess, the limiting reactant is lead(II) nitrate (Pb(NO₃)₂).
<u><em>Using cross multiplication:</em></u>
1.0 mole of Pb(NO₃)₂ produces → 1.0 mole of PbCl₂, from the stichiometry.
∴ 2.63 mole of Pb(NO₃)₂ produces → 2.63 mole of PbCl₂.
- The mass of PbCl₂ produced (the calculated yield) = n x molar mass = (2.63 mol) (278.1 g/mol) = 730.52 g.
∴ The percent yield % of lead(II) chloride (PbCl₂) = [(actual yield) / (calculated yield)] x 100 = [(650 g) / (730.52)] x 100 = 88.98 %.
Answer:
V = 65.81 L
Explanation:
En este caso, debemos usar la expresión para los gases ideales, la cual es la siguiente:
PV = nRT (1)
Donde:
P: Presion (atm)
V: Volumen (L)
n: moles
R: constante de gases (0.082 L atm / mol K)
T: Temperatura (K)
De ahí, despejando el volumen tenemos:
V = nRT / P (2)
Sin embargo como estamos hablando de condiciones normales de temperatura y presión, significa que estamos trabajando a 0° C (o 273 K) y 1 atm de presión. Lo que debemos hacer primero, es calcular los moles que hay en 50 g de amoníaco, usando su masa molar de 17 g/mol:
n = 50 / 17 = 2.94 moles
Con estos moles, reemplazamos en la expresión (2) y calculamos el volumen:
V = 2.94 * 0.082 * 273 / 1
<h2>
V = 65.81 L</h2>
The reaction between hydrogen (H2) and fluorine (F2) is given below,
H2 + F2 ---> 2HF
One mole of both hydrogen and fluorine yields to 2 moles of hydrogen fluoride. This can also be expressed as, 2 grams of hydrogen and 38 grams of fluorine will form 40 grams of hydrogen fluoride. From the given, only 20 grams of HF is formed with 19 g of it being fluorine. Thus, the percentage fluorine of the compound formed is 95%.
Number 2 lower entropy and higher entropy