Question

Consider the reaction.

Answer 1

Answer:

C.62.5 g

Explanation:

At STP,  

Pressure = 1 atm  

Temperature = 273.15 K

Given, Volume = 100.0 L

Using ideal gas equation as:

$PV=nRT$

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

1 atm × 100.0 L = n ×0.0821 L atm/ K mol  × 273.15 K  

⇒n = 4.4592 moles

From the given reaction,

$N_2_{(g)}+3H_2_{(g)}\rightarrow 2NH_3_{(g)}$

2 moles of ammonia is formed when 1 mole of nitrogen gas undergoes reaction.

Also, 1 mole of ammonia is formed when 1/2 mole of nitrogen gas undergoes reaction.

4.0446 moles of ammonia is formed when 1/2*4.4592 moles of nitrogen gas undergoes reaction.

Moles of nitrogen = 2.2296 moles

Molar mass of nitrogen, $N_2$ = 28.0134 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$2.2296\ moles= \frac{Mass}{28.0134\ g/mol}$

Mass = 62.5 g

Answer 2

Answer:The correct answer is option C.

Explanation:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

At STP, 1 mol of gas occupies 22.4 L of volume

So, 100.0 L volume of $NH_3$will be occupied by :

$\frac{1}{22.4 L}\times 100.0 L=4.4642 moles$

According to reaction 2 moles $NH_3$ are obtained from 1 mole of $N_2$.

Then,4.4642 moles of $NH_3$ will be obtained from :

$\frac{1}{2}\times 4.4642$ that is 2.2321 moles of $N_2$

Mass of $N_2$ gas:

Moles of $N_2$ gas  Molar mass of $N_2$ gas:

$2.2321 mol\times 28 g/mol=62.4988\approx 62.50 g$

Hence, the correct answer is option C.