Th actual yield of the reaction is 24.86 g
We'll begin by calculating the theoretical yield of the reaction.
2Na + Cl₂ → 2NaCl
Molar mass of Na = 23 g/mol
Mass of Na from the balanced equation = 2 × 23 = 46 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl from the balanced = 2 × 58.5 = 117 g
From the balanced equation above,
46 g of Na reacted to produce 117 g of NaCl.
Therefore,
11.5 g of Na will react to produce = (11.5 × 117) / 46 = 29.25 g of NaCl.
Thus, the theoretical yield of NaCl is 29.25 g.
Finally, we shall determine the actual yield of NaCl.
- Theoretical yield = 29.25 g
Actual yield = Percent yield × Theoretical yield
Actual yield = 85% × 29.25
Actual yield = 0.85 × 29.25 g
Actual yield = 24.86 g
Learn more about stoichiometry: brainly.com/question/25899385
This can be done through electrolysis. Electrolysis is the separation of a substance into two or more substances that may differ from each other and from the original substance by passing an electric current through a solution that contains ions.
In the case of copper, we use a copper (II) sulphate solution which we put in a large beaker. The impure copper will be used as the positive electrode (anode) and for the negative electrode (cathode) will be a bar of pure copper.
When the electric current is switched on, the bar of pure copper which is the cathode increases greatly in size as copper ions leave the anode of impure copper and attach to the cathode. The anode becomes smaller and smaller as it loses copper ions until all that is left of it is impurities in form of a sludge beneath it.
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
Answer:
Mass of aluminium in sample = 3.591 g ≅ 3.6 grams
Explanation:
Given that, A sample of aluminum absorbs 50.1 J of heat, upon which the temperature of the sample increases from 20.0°C to 35.5°C.
the specific heat of aluminum is 0.900 J/g- °C
The relation between heat absorbed and change in temperature is given by, Q = msΔT.
where Q = heat absorbed
m = mass of the substance
s = specific heat of substance
ΔT = change in temperature
Now, in our case, Q = 50.1 J ; s = 0.900 J/g- °C; ΔT= 35.5-20 = 15.5°C
⇒ m = 
⇒ m = 
= 3.591 g ≅ 3.6 g
⇒ m ≅ 3.6 g
If you can a number of electrons in a atom you will get a ion of the element
