Stoichiometry numerical-
Chem4kids.com
Question- 23 gm of sodium reacts with water. Calculate
(i) the weight of H2 liberated
(ii) the moles of H2 liberated
(iii) the volume of H2 liberated at NTP
Solution –
(i) Reaction is,
2Na + 2 H2O —> 2NaOH + H2
2 x 23 = 46 gm 2×1= 2 gm
46 gm Na produces = 2 gm H2
23 gm Na produces = 2x 23 / 46 gm H2 = 1.0 gm H2
weight of H2 liberated= 1.0 gm
(ii) 2Na + 2 H2O —> 2NaOH + H2
2 mole 1 mole
moles of sodium = mass / atomic mass = 23 /23 = 1 mole
moles of sodium =1.0
2.0 mole Na produces = 1 mole H2
1.0 mole Na produces = 1 x1 / 2 gm H2 = 0.5 mole H2
Mole of H2 liberated= 0.5
(iii) Volume of 1 mole H2 at NTP = 22.4 L
Volume of 0.5 mole H2 at NTP = 22.4 x 0.5 L
= 11.2 L = 11200 ml
Volume of H2 liberated= 11.2 L = 11200 ml
Question 2) Calculate the weight of lime (CaO) obtained by heating 200 Kg of 95% pure lime stone (CaCO3).
Solution –
95% means,
100 Kg impure sample has pure CaCO3 = 95 Kg
200 Kg impure sample has pure CaCO3 = 95 x 200 / 100 Kg = 190 Kg
Heating of CaCO3 is given by the equation,
CaCO3 —> CaO + CO2
40+12+3×16 = 100 gm or 100Kg 40+16=56gm or 56 Kg CaO
100 Kg CaCO3 gives CaO = 56 Kg
190 Kg CaCO3 gives CaO = 56x 190 / 100 Kg= 106.4 Kg
Weight of lime (CaO) obtained= 106.4 Kg
Question 3) Calculate the mass of O2 required to completely burn 14 gm of ethylene.
Solution –
C2H4 +. 3O2 —> 2CO2 + 2H2O
1 mole. 3 mole 2 mole 2 mole
molar mass of ethylene = 12×2+ 4 x 1= 28 gm / mole
mass of ethylene= 14 gm
moles of ethylene= mass / molar mass = 14/28 = 0.5 mole
moles of ethylene= 0.5
1 mole ethylene requires O2 = 3 mole
0.5 mole ethylene requires O2 = 3 x 0.5 mole= 1.5 mole
moles of O2 = 1.5
molar mass of O2 = 16×2= 32 gm / mole
mass = mole x molar mass = 1.5 x 32 = 48 gm O2
Mass of O2 = 48 gm
Question 4) Calculate the weight of water ( gm) produced by the combustion of 32 gm methane.
Solution –
CH4(g) + 2O2(g) —> CO2(g) + 2H2O(g)
12+4×1=16 gm 2( 2×1+16) = 36 gm
16 gm CH4 produces H2O = 36 gm
32 gm CH4 produces H2O = 36 x 32 / 16 gm = 72 gm
Weight of water = 72 gm
