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=![\frac{2.0moles[tex]Na_{2}SO_{4}](https://tex.z-dn.net/?f=%5Cfrac%7B2.0moles%5Btex%5DNa_%7B2%7DSO_%7B4%7D)
}{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.
4. Molar mass of silver m Ms~=108 g/mol
Hence there are n=54*(1/108)=0.5 mols of Silver in 54 grams of Silver.
5. 6.3*(108/1)=680.4g
6. Avogadro's number : Na~=6.022×10^23<span>. </span>
6.0*(6.022*10^23/1)=36.132*10^23 atoms
7. Molar mass of Krypton : Mk=84 g/mol
112/84=1.33 moles of Kr
8. 1.93*10^24*(1/(6.022×10^23))=3.2 moles KF
9. Molar mass of Silicon : Ms=28 g/mol
86.2*(1/28)*(6.022×10^23/1)=18.5*10^23 atoms of silicon
10. Molar mass of Magnesium : M1=24 g/mol
4.8*10^24*(1/(6.022×10^23))*(24/1)=191 g Mg
The formula of butane is C4H10 but I don't how many atoms it contains though
To solve this problem, the dilution equation (M1 x V1 = M2 X V2) must be used. The given values in the problem are M1= 12.0 M, V1= 30 mL, and M2= 0.160 M. To solve for V2,
V2=M1xV1/M2
V2= (12x30)/0.16
V2= 2,250 mL.
The correct answer is 2.25 L.
