Answer:
Silver Acetate would be the Limiting Reagent.
Explanation:
The balance chemical equation for the given double displacement reaction is as;
HCl + AgC₂H₃O₂ → AgCl + HC₂H₃O₂
Step 1: <u>Calculate Moles of Starting Materials:</u>
Moles of HCl:
Moles = Mass / M.Mass
Moles = 72.9 g / 36.46
Moles = 1.99 moles
Moles of AgC₂H₃O₂:
Moles = 150 g / 166.91 g/mol
Moles = 0.898 moles
Step 2: <u>Find out Limiting reagent as:</u>
According to balance chemical equation.
1 mole of HCl reacts with = 1 mole of AgC₂H₃O₂
So,
1.99 moles of HCl will react with = X moles of AgC₂H₃O₂
Solving for X,
X = 1.99 mol × 1 mol / 1 mol
X = 1.99 mol of AgC₂H₃O₂
Hence, to completely consume 1.99 moles of Hydrochloric acid we will require 1.99 moles of Silver Acetate, But, we are provided with only 0.898 moles of Silver Acetate. This means Silver Acetate will consume first in the reaction therefore, it is the LIMITING REAGENT.
Answer:
Methanol would be used as a reagent in excess, since it is a very low-cost solvent. For product isolation, the first thing to do is remove the methanol through a distillation process. The residue produced can be dissolved in diethyl ether. Using a NaHCO₃ solution, extraction is performed. When it separates into two phases, the product will be in the ether and the reagent in the aqueous phase. The ether can also be removed by distillation, and at the end of this process you will have the product you want.
Explanation:
The law of conservation of mass applies to every reaction. In this case, you start with 1 Mg, 2 H, and 2CL and end up with the same five only their bonds have been rearranged, or in other words, they are joined up differently.
Gay-Lussac's law gives the relationship between pressure and temperature of gas. For a fixed amount of gas, pressure is directly proportional to temperature at constant volume.
P/T = k
where P - pressure , T - temperature and k - constant
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting the values in the equation
T = 4342 K
initial temperature was 4342 K
Conversion of mole to grams
k in mole = 1 mole/ atomic mass
K in mole =1/ 39.0983 g/mole
= 0.255765 g/mole
converting 40 grams of K
K 40 grams x [ 1 mole/ 39.0983 grams] = 1.0230623 mole
There are 1.0230623 moles of K in 40 K of Potassium
