What will be mass of $89.6 L$ of ammonia $(N H_{3})$ gas at $S T P$ ?

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Solution

$89.6 L$ of $N H_{3}$ gas at $S T P = \frac{89.6}{22.4} = 4 m o l$
$G M M$ of $N H_{3} = 14 + 3 = 17 g$
Mass $= m o l e \times G M M = 4 \times 17 = 68 g$

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Similar questions

(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?

Q. Statement 1 : Ammonia gas,

N H_{3}

has a smaller density than argon gas,

A r

, at

S T P

.
Statement 2: The density of a gas at

S T P

is found by dividing the molar mass by

22.4

litres.

Q. The mass of

112 {c m}^{3}

N H_{3}

gas at STP is:

Q. What will be the mass of

224 c m^{3}

of nitrogen(

N_{2}

) gas at STP

What is the mass of $22.4 {dm}^{3}$ of the following gas at STP?

(a) ammonia

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What will be mass of 89.6L of ammonia (NH3) gas at STP?

89.6L of NH3 gas at STP=89.622.4=4molGMM of NH3=14+3=17gMass =mole×GMM=4×17=68g

What will be mass of $89.6 L$ of ammonia $(N H_{3})$ gas at $S T P$ ?

$89.6 L$ of $N H_{3}$ gas at $S T P = \frac{89.6}{22.4} = 4 m o l$
$G M M$ of $N H_{3} = 14 + 3 = 17 g$
Mass $= m o l e \times G M M = 4 \times 17 = 68 g$