Question

how many moles of hydrogen gas are contained in 21.3 l at stp?
g : p = 1.00 atm
t = 273 k
v = 21.3 l
u : n = ? mol
e : pv = nrt

• use r = 0.0821 \frac{l\cdot atm}{mol\cdot k}

moles (n)

Explanation:

Step1: Rearrange ideal gas law for n

$n = \frac{PV}{RT}$

Step2: Substitute values into formula

$n = \frac{(1.00\ \text{atm})(21.3\ \text{L})}{(0.0821\ \frac{\text{L·atm}}{\text{mol·K}})(273\ \text{K})}$

Step3: Calculate the result

$n = \frac{21.3}{0.0821 \times 273} \approx 0.951$

Answer:

0.951 mol