A Calculate the mass fo ammonia that can be obtained from 21-4 g of NH4Cl by the reaction- 2NH4Cl - Ca OH2 - CaCl2 - 2H2O - 2NH3 b what will be ht volume of ammonia when measured at STP-

(a) Molar mass = 53.5g/mol Therefore No of moles in 21.4g of NH_4Cl = 21.4 / 53.5 = 0.4 mol (b) At STP 1.0 mol has volume = 22.4L The volume of 0.4 mol = ...

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Byju's Answer

Standard X

Chemistry

Mole Concept and Avogadro's Number

a Calculate t...

Question

(a) Calculate the mass fo ammonia that can be obtained from 21.4 g of $N H_{4} C l$ by the reaction:
$2 N H_{4} C l + C a (O H)_{2} \to C a C l_{2} + 2 H_{2} O + 2 N H_{3}$
(b) what will be ht volume of ammonia when measured at STP?

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Solution

(a) Molar mass = 53.5g/mol
Therefore No of moles in 21.4g of $N H_{4} C l = \frac{21.4}{53.5}$ = 0.4 mol
(b) At STP 1.0 mol has volume = 22.4L
The volume of 0.4 mol = 0.422.4 = 8.96L

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Similar questions

Q. How many grams of ammonia is obtained when 32.6 g of ammonium chloride reacts with calcium hydroxide during the laboratory preparation of ammonia?

[2 N H_{4} C l + C a (O H)_{2} \to C a C l_{2} + 2 H_{2} O + 2 N H_{3}]

[N=14, H=1, O=16, S=32, Cl=35.5]

IN the synthesis of sodium carbonate, the recovery of ammonia is done by treating $N H_{4} C l$ with $C a (O H)_{2}$ The by product obtained in this process is
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(a) Calculate the mass fo ammonia that can be obtained from 21.4 g of NH4Cl by the reaction: 2NH4Cl+Ca(OH)2→CaCl2+2H2O+2NH3 (b) what will be ht volume of ammonia when measured at STP?

(a) Molar mass = 53.5g/mol Therefore No of moles in 21.4g of NH4Cl=21.453.5 = 0.4 mol (b) At STP 1.0 mol has volume = 22.4L The volume of 0.4 mol = 0.422.4 = 8.96L

(a) Calculate the mass fo ammonia that can be obtained from 21.4 g of $N H_{4} C l$ by the reaction:
$2 N H_{4} C l + C a (O H)_{2} \to C a C l_{2} + 2 H_{2} O + 2 N H_{3}$
(b) what will be ht volume of ammonia when measured at STP?

(a) Molar mass = 53.5g/mol
Therefore No of moles in 21.4g of $N H_{4} C l = \frac{21.4}{53.5}$ = 0.4 mol
(b) At STP 1.0 mol has volume = 22.4L
The volume of 0.4 mol = 0.422.4 = 8.96L