Stoichiometry numerical-

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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= 2gm

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 = 22.4x 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 stona (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

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 = 3x 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