What is the volume occupied by 3.00 mol of hydrogen gas at STP? *

Chemistry

1 answer:

xxTIMURxx [149]2 years ago

6 0

Answer: V=67.2 L

Explanation:

For this problem we will need to use the Ideal Gas Law.

Ideal Gas Law: PV=nRT

P=1.00 atm (STP)

V=?

n=3.00 mol

R=0.08206Latm/Kmol

T=273.15 K (STP)

To find V, we would manipulate the equation to V=nRT/P

$V=\frac{(3.00 mol)(0.08206Latm/Kmol)(273.15K)}{1.00 atm}$

$V=67.244 L$

With significan figures, our answer is V=67.2 L.

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