Question

What is the volume occupied by 51.0 g of ammonia gas (nh3) at stp?

Answer 1

The volume that is occupied by 51.0 grams of ammonia gas(NH3) at STP is calculated as follows

find the moles of NH3= mass/molar mass

= 51.0 g/17 g/mol = 3 moles

At STP 1mole = 22.4 liters, what about 3 moles of NH3

= 3moles /1 mole x 22.4 L =67.2 liters will occupied by 51.0 g of NH3

Answer 2

Answer : The volume of ammonia gas at STP is, 67.2 liters

Explanation : Given,

Mass of ammonia = 51.0 g

Molar mass of ammonia = 17 g/mole

As we know that,

At STP, 1 mole of gas contains 22.4 L volume of gas.

First we have to calculate the moles of ammonia gas.

`\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}=(51.0g)/(17g/mole)=3mole`

Now we have to calculate the volume of ammonia gas at STP.

As, 1 mole of ammonia gas contains 22.4 L volume of ammonia gas

So, 3 moles of ammonia gas contains
`3* 22.4L=67.2L` volume of ammonia gas

Therefore, the volume of ammonia gas at STP is, 67.2 liters