Answer:
a) The theoretical yield is 408.45g of 
b) Percent yield = 
Explanation:
1. First determine the numer of moles of
and
.
Molarity is expressed as:
M=
- For the 
M=
Therefore there are 1.75 moles of 
- For the 
M=
}{1Lsolution}[/tex]
Therefore there are 2.0 moles of 
2. Write the balanced chemical equation for the synthesis of the barium white pigment,
:

3. Determine the limiting reagent.
To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:
- For the
:

- For the
:

As the
is the smalles quantity, this is the limiting reagent.
4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:
Percent yield = 
Percent yield = 
The real yield is the quantity of barium white pigment you obtained in the laboratory.
I choose the option A.
The electron absorbs energy from specific wavelength then moving from a lower energy orbital to a higher energy orbital.
Answer:
Mass of carbon dioxide produced = 52.8 g
Explanation:
Given data:
Mass of carbon react = 14.4 g
Mass of oxygen = 56.5 g
Mass of oxygen left = 18.1 g
Mass of carbon dioxide produced = ?
Solution:
C + O₂ → CO₂
Number of moles of C:
Number of moles = mass/ molar mass
Number of moles = 14.4 g/ 12 g/mol
Number of moles = 1.2 mol
18.1 g of oxygen left it means carbon is limiting reactant.
Now we will compare the moles of C with CO₂.
C : CO₂
1 : 1
1.2 : 1.2
Mass of CO₂:
Mass = number of moles × molar mass
Mass = 1.2 mol × 44 g/mol
Mass = 52.8 g