Answer:
Here's what I get
Explanation:
1. Names
I. CH₃-CH₂-COOH = 49. propanoic acid
II. CH₃-CH₂-OH = 46. ethanol
III. CH₃-COO-CH₂-CH₂-CH₃ = 47. propyl ethanoate
IV. H-O-CH₂-CH₂-CH₃ = 48. propan-1-ol
V. H-COO-CH₃ = 51. methyl methanoate
VI. CH₃-COOH = 50. ethanoic acid
2. Precursors
52. methyl propionate ⇒ methanol + propanoic acid
53. ethyl methanoate ⇒ ethanol + methanoic acid
Explanation:
The degree of unsaturation is given as;
Degree of unsaturation = (2C + 2 + N - H - X) / 2
(a) C5H6
Using the formular above;
Degree of unsaturation = [ 2(5) + 2 - 6] / 2
Degree of unsaturation = [ 12 - 6 ] / 2 = 3
(b) C10H6Cl2
Using the formular above;
Degree of unsaturation = [ 2(10) + 2 - 6 -2 ] / 2
Degree of unsaturation = [ 22 - 8 ] / 2 = 7
(c) C4H3NO
Using the formular above;
Degree of unsaturation = [ 2(4) + 2 - 3 -1 ] / 2
Degree of unsaturation = [ 10 - 4 ] / 2 = 3
Note: Oxygen is ignored because its presence has no effect on the degree of unsaturation.
The amount of hydrogen chloride that can be made is 1064 g
Why?
The two reactions are:
2H₂O → 2H₂ + O₂ 75.3 % yield
H₂ + Cl₂ → 2HCl 69.8% yield
We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!
Have a nice day!
#LearnwithBrainly
NaOH + CH3COOH -> CH3COONa + H20
Balanced equation:
<span>CaO + 2 HCl --> CaCl2 + H2O </span>
<span>Calculate moles of each reactant: </span>
<span>60.4 g CaO / 56.08 g/mol = 1.08 mol CaO </span>
<span>69.0 g HCl / 36.46 g/mol = 1.89 mol HCl </span>
<span>Identify the limiting reactant: </span>
<span>Moles CaO needed to react with all HCl: </span>
<span>1.89 mol HCl X (1 mol CaO / 2 mol HCl) = 0.946 mol CaO </span>
<span>Because you have more CaO than that available, HCl is the limiting reactant. </span>
<span>Calculate moles and mass CaCl2: </span>
<span>1.89 mol HCl X (1 mol CaCl2 / 2mol HCl) X 111.0 g/mol = 105 g CaCl2</span>
