StudentAidUnit 15: Stoichiometry — Numericals
10th Class Chemistry · Unit 15: Stoichiometry
1.Calculate the number of moles, volume and number of molecules in 4.0 g of CH₄ at RTP.
Given
Mass of CH₄
m = 4.0 g
Molar mass of CH₄
M = 12 + (1 × 4) = 16 g/mol
Formula
Number of moles
n = mM
Working
Substituting
n = 4.016 = 0.25 mol
Result
Number of moles
n = 0.25 mol
Formula
Volume at RTP
V = n × Molar volume
Working
Substituting
V = 0.25 × 24 = 6 dm3
Result
Volume
V = 6 dm3
Formula
Number of molecules
N = n × NA
Working
Substituting
N = 0.25 × 6.02 × 1023
Result
Number of molecules
N = 1.5 × 1023
2.What mass of NaCl need to be dissolved in 150 cm³ of a solution if the concentration of the solution is 0.4 mol/dm³?
Given
Volume of solution
V = 150 cm3
Concentration
c = 0.4 mol/dm3
Molar mass of NaCl
M = 23 + 35.5 = 58.5 g/mol
Convert volume to dm³
V = 1501000 = 0.15 dm3
Formula
Number of moles from concentration
n = c × V
Working
Substituting
n = 0.4 × 0.15 = 0.06 mol
Formula
Mass of solute
m = n × M
Working
Substituting
m = 0.06 × 58.5
Result
Mass of NaCl
m = 3.51 g
3.Find out the percentage of sodium in NaHCO₃.
Given
Compound
NaHCO3
Molar mass of NaHCO₃
M = 23 + 1 + 12 + (16 × 3) = 84 g/mol
Formula
Percentage of sodium
%Na = frac{Atomic mass of Na{Molar mass of compound × 100
Working
Substituting
%Na = 2384 × 100
Result
Percentage of sodium
%Na = 27.38%
4.A sulphide of iron contains 1.926 g of sulphur and 2.333 g of iron. Find out the empirical formula of this compound.
Given
Mass of Sulphur
mS = 1.926 g
Mass of Iron
mFe = 2.333 g
Moles of Sulphur
nS = 1.92632 = 0.06 mol
Moles of Iron
nFe = 2.33356 = 0.04 mol
Divide by least value
S = 0.060.04 = 1.5, quad Fe = 0.040.04 = 1
Multiply by 2 to get whole numbers
Fe:S = 2:3
Result
Empirical formula
Fe2S3
5.A compound contains by mass, 40.0% C, 6.71% H and 53.3% oxygen. 0.320 mole of this compound weighs 28.8 g. What is the molecular formula of this compound?
Given
Percentage of C
%C = 40.0%
Percentage of O
%O = 53.3%
Moles of compound
n = 0.320 mol
Mass of compound
m = 28.8 g
Step 1: Moles of C in 100 g
nC = 4012 = 3.33
Moles of H in 100 g
nH = 6.711.008 = 6.65
Moles of O in 100 g
nO = 53.316 = 3.33
Step 2: Atomic ratio
C:H:O = 3.33:6.65:3.33 = 1:2:1
Result
Empirical formula
CH2O
Empirical formula mass
Memp = 12+2+16 = 30
Molar mass of compound
M = 28.80.320 = 90
n value
n = frac{M}{Memp = 9030 = 3
Molecular formula
C3H6O3
6.The formation of ammonia gas is given by the equation 3H₂(g) + N₂(g) → 2NH₃(g). If you start reacting 12 g of H₂ with 64 g of N₂, which will be the limiting reactant?
Given
Equation
3H2(g) + N2(g) to 2NH3(g)
Mass of Hydrogen
mH2 = 12 g
Mass of Nitrogen
mN2 = 64 g
Moles of H₂
nH2 = 122.016 = 5.95 mol
Moles of N₂
nN2 = 6428 = 2.28 mol
From equation: 3 mol H₂ produces 2 mol NH₃
5.95 mol H2 to 23 × 5.95 = 4 mol NH3
From equation: 1 mol N₂ produces 2 mol NH₃
2.28 mol N2 to 2 × 2.28 = 4.56 mol NH3
Result
Limiting reactant
H2 is the limiting reactant
7.When 305 g of AgNO₃ is reacted with excess of MgCl₂ it produces 23.7 g of Mg(NO₃)₂. What is the percentage yield of the reaction?
Given
Equation
2AgNO3(aq) + MgCl2(aq) to Mg(NO3)2(aq) + 2AgCl(l)
Mass of AgNO₃
m = 305 g
Actual yield of Mg(NO₃)₂
mactual = 23.7 g
Molar mass of AgNO₃
M = 108 + 14 + 48 = 170 g/mol
Moles of AgNO₃
n = 305170 = 1.794 mol
From equation: 2 mol AgNO₃ gives 1 mol Mg(NO₃)₂
1.80 mol AgNO3 to 12 × 1.794 = 0.897 mol Mg(NO3)2
Molar mass of Mg(NO₃)₂
M = 24 + 14 × 2 + 48 × 2 = 148 g/mol
Theoretical yield
mtheo = 0.897 × 148 = 132.756 g
Formula
Percentage yield
%yield = frac{Actual yield{Theoretical yield × 100
Working
Substituting
%yield = 23.7132.756 × 100
Result
Percentage yield
%yield = 17.85%
8.A sample of metal has a total mass of 3.66 kg and contains 2.45 kg of gold. What is the percentage purity of gold in this sample?
Given
Total mass of sample
mtotal = 3.66 kg
Mass of gold
mgold = 2.45 kg
Formula
Percentage purity
%purity = frac{Mass of pure gold{Total mass of sample × 100
Working
Substituting
%purity = 2.453.66 × 100
Result
Percentage purity of gold
%purity = 66.94%