Answer:
Explanation:
Hello,
In this case, we write the reaction again:
In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:
Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:
But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:
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Answer:
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Explanation:
The reaction is
CaC₂(s) + 2H₂O (l) -----> Ca(OH)₂ (s) + C₂H₂ (g)
As we have data of gas ethyne (or acetylene), C₂H₂
We can calculate the moles of acetylene and from this we can estimate the mass of calcium carbide taken
the moles of acetylene will be calculated using ideal gas equation
PV =nRT
R = gas constant = 0.0821 Latm/molK
T = 385 K
V = volume = 550 L
P = Pressure = 1.25 atm
n = moles = ?
n = PV /RT = 1.25 X 550 / 0.0821 X 385 = 21.75 mol
As per balanced equation these moles of acetylene will be obtained from same moles of calcium carbide
moles of calcium carbide = 21.75mol
molar mass of CaC₂ = 40 + 24 = 64
mass of CaC₂ = moles X molar mass = 21.75 X 64 = 1392g
Answer:
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Answer:
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