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.

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

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 $3\times 22.4L=67.2L$ volume of ammonia gas

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