Question

0.817 grams of an unknown gas occupies 1.25 L at 27.0 C and a pressure of 765

Answer 1

The gas is most likely methane (CH₄)

Further explanation

In general, the gas equation can be written  

$\large{\boxed{\bold{PV=nRT}}}$

where  

P = pressure, atm  

V = volume, liter  

n = number of moles  

R = gas constant = 0.08205 l.atm / mol K  

T = temperature, Kelvin  

$\tt PV=\dfrac{mass}{M}RT\\\\M=\dfrac{mRT}{PV}=\dfrac{0.817\times 0.082\times 300}{1,00658\times 1.25}=15.97\approx=16$

So the gas is most likely methane (CH₄ )