Answer:
Limiting reagent = lead(II) nitrate
Theoretical yield = 3.75435 g
% yield = 65.26 %
Explanation:
Considering:
Or,
Given :
For potassium chloride :
Molarity = 1.20 M
Volume = 25.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 25.0×10⁻³ L
Thus, moles of potassium chloride :
<u>Moles of potassium chloride = 0.03 moles</u>
For lead(II) nitrate :
Molarity = 0.900 M
Volume = 15.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 25.0×10⁻³ L
Thus, moles of lead(II) nitrate :
<u>Moles of lead(II) nitrate = 0.0135 moles</u>
According to the given reaction:
2 moles of potassium chloride react with 1 mole of lead(II) nitrate
1 mole of potassium chloride react with 1/2 mole of lead(II) nitrate
0.03 moles potassium chloride react with 0.03/2 mole of lead(II) nitrate
Moles of lead(II) nitrate = 0.015 moles
<u>Limiting reagent is the one which is present in small amount. Thus, lead(II) nitrate is limiting reagent. (0.0135 < 0.015)</u>
The formation of the product is governed by the limiting reagent. So,
1 mole of lead(II) nitrate gives 1 mole of lead(II) chloride
0.0135 mole of lead(II) nitrate gives 0.0135 mole of lead(II) chloride
Molar mass of lead(II) chloride = 278.1 g/mol
Mass of lead(II) chloride = Moles × Molar mass = 0.0135 × 278.1 g = 3.75435 g
<u>Theoretical yield = 3.75435 g</u>
Given experimental yield = 2.45 g
<u>% yield = (Experimental yield / Theoretical yield) × 100 = (2.45/3.75435) × 100 = 65.26 %</u>
Answer:
1.6 L
Explanation:
Using Charle's law
Given ,
V₁ = 1.5 L
V₂ = ?
T₁ = 12 °C
T₂ = 32 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (12 + 273.15) K = 285.15 K
T₂ = (32 + 273.15) K = 305.15 K
Using above equation as:
New volume = 1.6 L