Question

A sample of Ammonia gas at 650mmHg and 15 degree has a mass of 56.8g.calculate the volume occupied by the gas. Take N=14, H=1,R=8.3145jk.

Answer 1

Answer:

The first thing we have to do is change and state all the units so that we can use our ideal gas law equation ($PV = nRT$).

650 mmHg is a pressure unit, we have to convert this to kiloPascals. We know that 760 mmHg gives us 101 kPa.

$650 \ mmHg \ * \ \frac{101kPa}{760 mmHg} = 86 \ kPa$

P = 86kPa

T = 15°C + 273K = 288K

R (Gas constant) = 8.31 kj/mol  

Molar mass of Ammonia ($NH_{3}$) = (1 x 3) + (14) = 17g/mol

n (moles) = $\frac{mass}{molar \ mass}$ $= \frac{56.8}{17} =$ 3.34 mol

V = ?

Rearrange the equation to solve for Volume:

$V = \frac{nRT}{P}$

Substitute the values inside:

V = $\frac{(3.34)(8.31)(288)}{(86)} = 93 L (rounded)$

Therefore 93 L of volume is occupied by the ammonia gas.