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<span> Mg(OH)2(s) + 2HCl(aq) yield MgCl2(aq) + 2H2O(l)
grams HCl required = (50.6 grams Mg(OH)2) * (1 mol Mg(OH)2 / 58.3197 grams Mg(OH)2) * (2 mol HCl / 1 mol Mg(OH)2) * (36.453 grams HCl / 1 mol HCl) = 63.26 grams HCl required
Since there are only 45.0 grams HCl, then HCl is the limiting reactant.
theoretical yield MgCl2 = (45.0 grams HCl) * (1 mol HCl / 36.453 grams HCl) * (1 mol MgCl2 / 2 mol HCl) * (95.211 grams MgCl2 / 1 mol MgCl2) = 58.6 grams MgCl2 </span>
Answer:
The answer to your question is: 83.9 %
Explanation:
Data
Cu = 31.8 g
S = 50 g
CuS = 40 g
yield = ?
Equation
Cu + S ⇒ CuS
MW Cu = 64 g
MW S = 32 g
MW CuS = 96 g
Ratio (theoretical/experimental)
Experimental 50/31.8 = 1.57
Theoretical 32/64 = 0.5 limiting reactant Cu
64 g of Cu ------------------ 96 g of CuS
31.8 g ------------------- x
x = (31.8 x 96) / 64
x = 47.7 g of CuS
% yield = (40/47.7) x 100
= 83.9 %
Complete Question
Question 096 Propose a three-step synthetic sequence to accomplish the transformation below.
Option 1 => 1) HBr, ROOR; 2) t-BuOK; 3) CH3CH2CCNa
Option 2 => 1) NaOEt; 2) HBr, ROOR; 3) CH3CH2CCNa
Option 3 => 1) t-BuOK; 2) NaNH2; 3) CH3CH2CCNa
Option 4 => 1) CH3CH2CH2CH2Br; 2) NaOEt; 3) HBr, ROOR
Option 5 => 1) t-BuOK; 2) HBr, ROOR; 3) CH3CH2CCNa
Option 6 => 1) NaOEt; 2) NBS, hν; 3) NaSBu
Answer:
The correct option is option 5
Explanation:
The mechanism of the reaction is shown on the first uploaded image
Ionized, become charged, become a cation
hopefully that helps
