Urea NH2CONH2 is formed when CO2 reacts with NH3- CO2- 2NH3- NH2CONH2-H2O NH3 and CO2 are obtained in the following steps- I- N2 - 3H22NH3 II- CaCO3CaO-CO2Amount of N2 and CaCO3 required to produce 60 kg of urea are X kg and Y kg respectively- The value of X-Y nearest integer value is -

The molar masses of CaCO_3, nbsp; N_2 and NH_2CONH_2 are 100 g/mol, 28 g/mol and 60 g/mol respectively. 1 mole of CaCO ...

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Byju's Answer

Standard XII

Chemistry

Stoichiometric Calculations

Urea NH2CON...

Question

Urea $(N H_{2} C O N H_{2})$ is formed when $C O_{2}$ reacts with $N H_{3}$ .
$C O_{2} + 2 N H_{3} ⟶ N H_{2} C O N H_{2} + H_{2} O$
$N H_{3}$ and $C O_{2}$ are obtained in the following steps:
$I$ : $N_{2} + 3 H_{2} 90 % - - \to 2 N H_{3}$
$I I$ : $C a C O_{3} 50 % - - \to C a O + C O_{2}$
Amount of $N_{2}$ and $C a C O_{3}$ required to produce $60$ kg of urea are $X$ kg and $Y$ kg respectively. The value of $X + Y$ (nearest integer value ) is :

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Solution

The molar masses of $C a C O_{3}, N_{2}$ and $N H_{2} C O N H_{2}$ are $100$ g/mol, $28$ g/mol and $60$ g/mol respectively.
$1$ mole of $C a C O_{3}$ and $1$ mole of $N_{2}$ gives $1$ mole of urea as per the stoichiometry.
Thus, $100$ kg of $C a C O_{3}$ and $28$ kg of $N_{2}$ gives $60$ kg of urea.
However, the yields are less than $100 %$ .
Hence, mass of nitrogen required $Y = \frac{28}{0.9} = 31$ kg.
Mass of $C a C O_{3}$ required $X = \frac{100}{0.50} = 200$ kg.
Hence, $X + Y = 200$ kg $+ 31$ kg $= 231$ kg.

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2NH3+CO2----------- NH2CONH2+H2O

Q. Calculate the change of entropy

Δ_{r} S_{m}^{\circ}

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N H_{3}

and

C O_{2}

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N H_{2} C O N H_{2}

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J K^{- 1} m o l^{- 1}

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N H_{3}

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80 % .

N H_{2} C O N H_{2} + 2 N a O H Δ - \to N a_{2} C O_{3} + 2 N H_{3}

6g of

N H_{2} C O N H_{2}

reacts with 8g

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x

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[Note: in g as nearest integer]

Q. Percent yield of

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80 %

N H_{2} C O N H_{2} + 2 N a O H Δ - \to N a_{2} C O_{3} + 2 N H_{3}

6

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How many covalent bonds are present in the following compounds?

(i) $C O_{2}$

(ii) $H_{2} O$

(iii) $C H_{4}$

(iv) $N_{2}$

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