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Chemistry

(i) A gas of mass 32 gms has a volume of 20 litres at S.T.P. Calculate the gram molecular weight of the gas.

(ii) How much Calcium oxide is formed when 82 g of calcium nitrate is heated? Also find the volume of nitrogen dioxide evolved:

2Ca(NO3)2 ⟶ 2CaO + 4NO2 + O2

[Ca = 40, N = 14, O = 16]

Stoichiometry

ICSE 2016

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Answer

(i) 20 lit. of gas weighs = 32 g.

1 mole = 22.4 lit. of gas will weigh = 3220\dfrac{32}{20} x 22.4 = 35.84 g.

Hence, gram mol. weight of the gas = 35.84 g.

(ii)

2Ca(NO3)2⟶2CaO+4NO2+O22[(40)+2(404[22.4]2(14+3(16))]+16) lit.=328 g112g\begin{matrix} 2\text{Ca}(\text{NO}3)2 & \longrightarrow & 2\text{CaO} & + & 4\text{NO}2 & + & \text{O}2 \ 2[(40) + & & 2(40 & & 4[22.4] \ 2(14 +3(16))] & & + 16) & & \text{ lit.} \ = 328 \text{ g} & & 112 \text{g} \ \end{matrix}

(a) 328 g of Ca(NO3)2 produces 112 g of calcium oxide.

∴ 82 g of Ca(NO3)2 will produce = 112328\dfrac{112}{328} x 82 = 27.99 g = 28 g

Hence, 28 g of calcium oxide is produced.

(b) 328 g of Ca(NO3)2 produces 4(22.4) lit of nitrogen dioxide.

∴ 82 g of Ca(NO3)2 will produce = 4×22.4328\dfrac{4 \times 22.4}{328} x 82 = 22.4 lit.

Hence, vol of nitrogen dioxide evolved = 22.4 lit.

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