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.
Answer:
The volume will decrease and the balloons will be smaller
Answer : The enthalpy of the reaction = -1839.6 KJ
Solution : Given,
= -520.0 KJ/mole
= -1699.8 KJ/mole
The balanced chemical reaction is,
Formula used :
We know that the standard enthalpy of formation of the element is equal to Zero.
Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.
Now, put all the values in above formula, we get
![\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]](https://tex.z-dn.net/?f=%5CDelta%20%28H_%7Bf%7D%29_%7Breaction%7D%3D%5B2moles%5Ctimes%20%28-1699.8%20KJ%2Fmole%29%7D%2B3moles%5Ctimes%20%280%5Ctext%7B%20KJ%2Fmole%7D%7D%29%5D-%5B%283moles%5Ctimes%28-520.0KJ%2Fmole%20%7D%2B4moles%5Ctimes%280%5Ctext%7B%20KJ%2Fmole%7D%29%5D)
= (-3399.6) + (1560)
= -1839.6 KJ
Answer:
The amount of energy liberated will be 49.38 J.
Explanation:
The amount of energy liberated (gibbs free energy) can be calculated using the following equation:
ΔG° = -nFε
n: amount of moles of electrons transfered
F: Faraday's constant
ε: cell potential
20.0 g of Zn is equal to 0.30 mol.
Two electrons are transfered during the reaction.
Therefore, n = 2x0.30 ∴ n = 0.60
ΔG° = - 0.60 x 96.485 x 0.853
ΔG° = 49.38 J
Answer:
211.63 g.
Explanation:
- Particles could refer to atoms, molecules, formula units.
- <em>Knowing that every one mole of a substance contains Avogadro's no. of molecules (NA = 6.022 x 10²³).</em>
<em><u>Using cross multiplication: </u></em>
1.0 mole → 6.022 x 10²³ molecules.
??? mole → 8.95 x 10²³ molecules.
- The no. of moles of magnesium acetate = (8.95 x 10²³ molecules) (1.0 mole) / (6.022 x 10²³ molecules) = 1.486 mol.
∴ The grams of magnesium acetate are in 8.95 x 10²³ formula units = n x molar mass = (1.486 mol)(142.394 g/mol) = 211.63 g.
