Q3. Sodium reacts with chlorine to form sodium chloride. In an experiment a stud
ID: 1030619 • Letter: Q
Question
Q3. Sodium reacts with chlorine to form sodium chloride. In an experiment a student reacted 1.50 g of sodium with 1.90 g of chlorine.
(a) Write the balanced equation for this reaction.
(b) Which is the limiting reactant? (Show calculations)
(c) Which is the excess reactant? How many grams of the excess reactant remains unreacted?
(d) How many grams of sodium chloride is expected to be produced during this reaction? (Show calculations)
(e) Suppose the percent yield of sodium chloride was reported to be 82% in this experiment, what must have been the actual yield of sodium chloride?
Explanation / Answer
a)
Balanced chemical equation is:
2 Na + Cl2 ---> 2 NaCl
b)
Molar mass of Na = 22.99 g/mol
mass(Na)= 1.5 g
use:
number of mol of Na,
n = mass of Na/molar mass of Na
=(1.5 g)/(22.99 g/mol)
= 6.525*10^-2 mol
Molar mass of Cl2 = 70.9 g/mol
mass(Cl2)= 1.9 g
use:
number of mol of Cl2,
n = mass of Cl2/molar mass of Cl2
=(1.9 g)/(70.9 g/mol)
= 2.68*10^-2 mol
Balanced chemical equation is:
2 Na + Cl2 ---> 2 NaCl
2 mol of Na reacts with 1 mol of Cl2
for 6.525*10^-2 mol of Na, 3.262*10^-2 mol of Cl2 is required
But we have 2.68*10^-2 mol of Cl2
so, Cl2 is limiting reagent
c)
Na is excess reactant
According to balanced equation
mol of Na reacted = (2/1)* moles of Cl2
= (2/1)*2.68*10^-2
= 5.36*10^-2 mol
mol of Na remaining = mol initially present - mol reacted
mol of Na remaining = 6.525*10^-2 - 5.36*10^-2
mol of Na remaining = 1.165*10^-2 mol
Molar mass of Na = 22.99 g/mol
use:
mass of Na,
m = number of mol * molar mass
= 1.165*10^-2 mol * 22.99 g/mol
= 0.2678 g
Answer: 0.268 g
d)
Molar mass of NaCl,
MM = 1*MM(Na) + 1*MM(Cl)
= 1*22.99 + 1*35.45
= 58.44 g/mol
According to balanced equation
mol of NaCl formed = (2/1)* moles of Cl2
= (2/1)*2.68*10^-2
= 5.36*10^-2 mol
use:
mass of NaCl = number of mol * molar mass
= 5.36*10^-2*58.44
= 3.132 g
Answer: 3.13 g
e)
% yield = actual mass*100/theoretical mass
82.0= actual mass*100/3.132
actual mass=2.568 g
Answer: 2.57 g
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