Answer:
The product is 1-chlorooctene
Explanation:
The alcoholic KOH cause the elimination reaction
It is <span>the attraction of an SO42– ion </span>
The volume of water that will be produced from the reaction will be 6.3 mL
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
The mole ratio of hydrogen sulfate to sodium hydroxide is 1:2.
Mole of hydrogen sulfate = 0.50 x 350/1000 = 0.175 moles
Mole of 15 grams sodium hydroxide = 15/40 = 0.375 moles
Thus, hydrogen sulfide is the limiting reagent.
Mole ratio of hydrogen sulfide to water = 1:2.
Equivalent mole of water = 0.175 x 2 = 0.35 moles
Mass of 0.35 moles of water = 0.35 x 18 = 6.3 grams.
1 gram of water = 1 ml.
Thus, 6.3 grams of water will be equivalent to 6.3 mL
More on stoichiometric calculation can be found here: brainly.com/question/27287858
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Pretty sure it’s C- hope this helps❤️
Answer:
66 g of CO₂
Solution:
The Balance Chemical Reaction is as follow,
C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O
Or,
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O ------- (1)
Step 1: Find out the limiting reagent as;
According to Equation 1,
56.1 g (2 mole) C₂H₂ reacts with = 160 g (5 moles) of O₂
So,
125 g of C₂H₂ will react with = X g of O₂
Solving for X,
X = (125 g × 160 g) ÷ 56.1 g
X = 356.5 g of O₂
It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.
Step 2: Calculate Amount of CO₂ produced as;
According to Equation 1,
160 g (5 mole) O₂ produces = 176 g (4 moles) of CO₂
So,
60.0 g of O₂ will produce = X g of CO₂
Solving for X,
X = (60.0 g × 176 g) ÷ 160 g
X = 66 g of CO₂
