WB Calculation
14
M r and A r are always defined as “…compared to 1/12 of a C-12 atom.”
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ws ans
Transcript of WB Calculation
Mr and Ar are always defined as
“…compared to 1/12 of a C-12 atom.”
Ar of C: 12 , Ar of H: 1
CnH2n+2 = (12)(n) + (2n+2)(1)30 = 12n + 2n + 228 = 14n n = 2
What is the value of n ?
same number of particles = same number of moles
e.g. 6 x 10 23 apples = 1 mole of apples
e.g. 6 x 10 23 books = 1 mole of books
(CO)
(Si)
(Sn)
(N2)
So you need to convert all the mass to mole and compare !!!
A 14 g of CO = 14 / 28 = 0.5 mol
B 14 g of C2H4 = 14 / 28 = 0.5 mol
C 14 g of Si = 14 / 28 = 0.5 mol
D 14 g of Sn = 14 / 119 = 0.12 mol
(CO)
(Si)
(Sn)
(N2)
14 g of N2
= 14 / 28 = 0.5 mol
(A) should be 2 moles CO react with 1 mole O2
(B) should be 2 moles CO2 obtained from 1 mole O2
(C) 28 kg CO = 28000 / 28 = 1000 mol
44 kg CO2 = 44000 / 44 = 1000 mol
���� this match the mole ratio of 2CO : 2CO2
Mole ratio of 2 : 1 : 2
Moles of CaCO3
= mass / Mr
= 10 / 100
= 0.1 mol
Step 1:
Mole ratio of 1 : 1 : 1
Moles of CaCO3
= mass / Mr
= 10 / 100
= 0.1 mol
CaCO3 : CO2
1 mol : 1 mol
0.1 mol : 0.1 mol
Step 1:
Step 2:
Mole ratio of 1 : 1 : 1
Moles of CaCO3
= mass / Mr
= 10 / 100
= 0.1 mol
CaCO3 : CO2
1 mol : 1 mol
0.1 mol : 0.1 mol
Step 1:
Step 2:
Vol = mol x 24
= 0.1 x 24
= 2.4 dm3
Step 3:
Mole ratio of 1 : 1 : 1
ANS : (D)
CH4 (g) + 2O2 (g) ���� CO2 (g) + 2H2O (l)
What volume of gas would be formed at r.t.p. when
20 cm3 of methane burns in 40 cm3 of oxygen ?
CH4 (g) + 2O2 (g) ���� CO2 (g) + 2H2O (l)
1 mol : 2 mol : 1 mol : 2 mol
20 cm3 : 40 cm3 : 20 cm3
You can only do this for gases !!!
ANS : (A)
7. A student reacted a solution of sodium hydroxide with an unknown acid. He found that 20.0 cm3 of 1.5 mol/dm3 of
sodium hydroxide exactly reacted with 10.0 cm3 of
1.0 mol/dm3 of the acid.
How many moles of sodium hydroxide would react with one mole of the acid?
Since the question is asking about moles, lets convert all the information to moles.
What formula can we use to find moles
involving solution?
7. A student reacted a solution of sodium hydroxide with an unknown acid. He found that 20.0 cm3 of 1.5 mol/dm3 of
sodium hydroxide exactly reacted with 10.0 cm3 of
1.0 mol/dm3 of the acid.
How many moles of sodium hydroxide would react with one mole of the acid?
Since the question is asking about moles, lets convert all the information to moles.
concentration = moles / volume
moles = conc. x volume
7. A student reacted a solution of sodium hydroxide with an unknown acid. He found that 20.0 cm3 of 1.5 mol/dm3 of
sodium hydroxide exactly reacted with 10.0 cm3 of
1.0 mol/dm3 of the acid.
How many moles of sodium hydroxide would react with one mole of the acid?
Mole of sodium hydroxide used = conc. x vol= 1.5 x 20/1000= 0.03 mol
7. A student reacted a solution of sodium hydroxide with an unknown acid. He found that 20.0 cm3 of 1.5 mol/dm3 of
sodium hydroxide exactly reacted with 10.0 cm3 of
1.0 mol/dm3 of the acid.
How many moles of sodium hydroxide would react with one mole of the acid?
Mole of sodium hydroxide used = conc. x vol= 1.5 x 20/1000= 0.03 mol
Mole of acid used = conc. x vol= 1.0 x 10/1000= 0.01 mol
7. A student reacted a solution of sodium hydroxide with an unknown acid. He found that 20.0 cm3 of 1.5 mol/dm3 of
sodium hydroxide exactly reacted with 10.00 cm3 of
1.0 mol/dm3 of the acid.
How many moles of sodium hydroxide would react with one mole of the acid?
moles of acid : moles of sodium hydroxide0.01 mol : 0.03 mol
1 mol : 3 mol
Therefore:
ANS : (B)
