Answer: The number of grams of [tex]H_2[/tex] in 1620 mL is 1.44 g
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 1 atm (at STP)
V = Volume of gas = 1620 ml = 1.62 L (1L=1000ml)
n = number of moles = ?
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]273K[/tex]
[tex]n=\frac{PV}{RT}[/tex]
[tex]n=\frac{1atm\times 16.2L}{0.0821Latm/K mol\times 273K}=0.72moles[/tex]
Mass of hydrogen =[tex]moles\times {\text {Molar mass}}=0.72mol\times 2g/mol=1.44g[/tex]
The number of grams of [tex]H_2[/tex] in 1620 mL is 1.44 g
