(a) Calculate the mass of nitrogen supplied to the soil by 5 kg of urea $[C O (N H_{2})_{2}]$
$[O = 16 : N = 14; C = 12; H = 1]$
(b) Calculate the volume occupied by 320 g of sulphur dioxide at STP $[S = 32; O = 16]$

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Solution

(a) 60 g urea has mass of nitrogen(N₂) = 28 g
So, 5000 g urea will have mass $= 28 \times \frac{5000}{60} = 2.33 k g$

(b) 64 g has volume = 22.4 litre
So, 320 g will have volume $= 22.4 \times \frac{320}{64} = 112 l i t r e s$

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Similar questions

What is the mass of nitrogen in 1000 kg of urea [CO(NH₂)₂]?
(H = 1; C = 12; N = 14; O = 16)

(i) Calculate the volume of 320 g of SO₂ at stp (Atomic mass: S = 32 and O = 16)
(ii) State Gay-Lussac's Law of combining volumes.
(iii) Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C₃H₈). (Atomic mass: C = 14, O = 16, H = 1, Molar Volume = 22.4 dm³ at stp)

Q. Ammonia reacts with sulphuric acid to give the fertilizer ammonium sulphate. Calculate the volume of ammonia [at STP] used to form

59 g

of ammonium sulphate.

[N = 14, H = 1, S = 32, O = 16]

(a)

(i) Calculate the volume of 320 of $S O_{2}$ at STP. (Atomic mass : S = 32 and O = 16).
(ii) State Gay-Lussac's law of combining volume
(iii) Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane $(C_{3} H_{8})$ (Atomic mass; C = 12, 0 = 16, H = 1, Molar volume $- 22.4 d m^{3}$

(b)

(i) An organic compound with vapour density = 94 contains
C = 12.67% H = 2.13 % and Br = 85.11% Find the molecular formula.
[Atomic mass ; C = 12, H = 1, Br = 80]
(ii) Calculate the mass of
(1) $10^{22}$ Atoms of sulphur.
(2) 0.1 mole of carbon dioxide.
[Atomic mass; S = 32, C = 12 and O = 16 and Avogadro's number $= 6 \times 10^{23}$

(a) A flask contains 3.2 g of sulphur dioxide. Calculate the following:
(i) The moles of sulphur dioxide present in the flask.
(ii) The number of molecules of sulphur dioxide present in the flask.
(iii) The volume occupied by 3.2 g of sulphur dioxide at STP. (S = 32, O = 16)

(b) The reaction of potassium permanganate (VII) with acidified iron (II) sulphate is given below:
2KMnO₄ + 10FeSO₄ + 8H₂SO₄ $\overset{}{\to}$ K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O
If 15.8 g of potassium permanganate (VII) was used in the reaction, calculate the mass of iron (II) sulphate used in the above reaction. (K = 39, Mn = 55, Fe = 56, S = 32, O = 16)

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(a) Calculate the mass of nitrogen supplied to the soil by 5 kg of urea [CO(NH2)2] [O=16:N=14;C=12;H=1] (b) Calculate the volume occupied by 320 g of sulphur dioxide at STP [S=32;O=16]

(a) 60 g urea has mass of nitrogen(N2) = 28 g So, 5000 g urea will have mass =28×500060=2.33 kg (b) 64 g has volume = 22.4 litre So, 320 g will have volume =22.4×32064=112 litres

(a) Calculate the mass of nitrogen supplied to the soil by 5 kg of urea $[C O (N H_{2})_{2}]$
$[O = 16 : N = 14; C = 12; H = 1]$
(b) Calculate the volume occupied by 320 g of sulphur dioxide at STP $[S = 32; O = 16]$

(a) 60 g urea has mass of nitrogen(N₂) = 28 g
So, 5000 g urea will have mass $= 28 \times \frac{5000}{60} = 2.33 k g$

(b) 64 g has volume = 22.4 litre
So, 320 g will have volume $= 22.4 \times \frac{320}{64} = 112 l i t r e s$