765.75
................................
Answer:
181.39g of AlCl3 is produced
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
3CuCl2•2H2O + 2Al → 2AlCl3 + 6H2O + 3Cu
Next, we shall determine the mass of Al that reacted and the mass of AlCl3 produced from the balanced equation. This is illustrated below:
Molar mass of Al = 27g/mol
Mass of Al from the balanced equation = 2 x 27 = 54g
Molar mass of AlCl3 = 27 + (3x35.5) = 133.5g/mol
Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g
Summary:
From the balanced equation above,
54g of Al reacted to produce 267g of AlCl3.
Next, we shall determine the theoretical yield of AlCl3. This can be achieved as shown below:
From the balanced equation above,
54g of Al reacted to produce 267g of AlCl3.
Therefore, 54.81g of Al will react to produce = (54.81 x 267)/54 = 271.01g of AlCl3.
Therefore, the theoretical yield of AlCl3 is 271.01g.
Finally, we shall determine the actual yield of AlCl3 produced from the reaction.
This can be obtain as follow:
Percentage yield of AlCl3 = 66.93%
Theoretical yield of AlCl3 = 271.01g
Actual yield of AlCl3 =?
Percentage yield = Actual yield/Theoretical yield x 100
66.93% = Actual yield /271.01g
Actual yield = 66.93% x 271.01
Actual yield = 66.93/100 x 271.01g
Actual yield = 181.39g.
Therefore, 181.39g of AlCl3 is produced from the reaction.
Density is the ratio of mass to the volume.
The mathematical expression is given as:
Now, density of isooctane =
Volume =
Since, 1 gallon = 3.78 L
So, 3.8 gal =
=
As, 1 L = 1000 mL
Therefore, =
Volume in mL =
Put the values,
=
Hence, mass of 3.8 gal of the gasoline is .
b. 26.0 g H2SO4 in 200.0 mL solution
c. 15.0 g NaCl dissolved to make 420.0 mL solution
Answer:
a) NaHCO3 = 0.504 M
b) H2SO4 = 1.325 M
c) NaCl = 0.610 M
Explanation:
Step 1: Data given
Moles = mass / molar mass
Molarity = moles / volume
a. 19.5 g NaHCO3 in 460.0 ml solution
Step 1: Data given
Mass NaHCO3 = 19.5 grams
Volume = 460.0 mL = 0.460 L
Molar mass NaHCO3 = 84.0 g/mol
Step 2: Calculate moles NaHCO3
Moles NaHCO3 = 19.5 grams / 84.0 g/mol
Moles NaHCO3 = 0.232 moles
Step 3: Calculate molarity
Molarity = 0.232 moles / 0.460 L
Molarity = 0.504 M
b. 26.0 g H2SO4 in 200.0 mL solution
Step 1: Data given
Mass H2SO4 = 26.0 grams
Volume = 200.0 mL = 0.200 L
Molar mass H2SO4 = 98.08 g/mol
Step 2: Calculate moles H2SO4
Moles H2SO4 = 26.0 grams / 98.08 g/mol
Moles H2SO4 = 0.265 moles
Step 3: Calculate molarity
Molarity = 0.265 moles / 0.200 L
Molarity =1.325 M
c. 15.0 g NaCl dissolved to make 420.0 mL solution
Step 1: Data given
Mass NaCl = 15.0 grams
Volume = 420.0 mL = 0.420 L
Molar mass NaCl = 58.44 g/mol
Step 2: Calculate moles NaCl
Moles NaCl = 15.0 grams / 58.44 g/mol
Moles NaCl = 0.256 moles
Step 3: Calculate molarity
Molarity = 0.256 moles / 0.420 L
Molarity =0.610 M
Answer:
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Explanation:
One second after the Big Bang, the universe was filled with neutrons, protons, electrons, anti-electrons, photons and neutrinos.Jun 17, 2017