Answer: The volume of product formed is 0.26 L
Explanation:
[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} H_2=\frac{0.02g}{2g/mol}=0.01moles[/tex]
[tex]2H_2(l)+O_2(l)\rightarrow 2H_2O(g)[/tex]
As [tex]O_2[/tex] is the the excess reagent, [tex]H_2[/tex] is the limiting reagent as it limits the formation of product.
According to stoichiometry :
2 moles of [tex]H_2[/tex] give = 2 moles of [tex]H_2O[/tex]
Thus 0.01 moles of [tex]H_2[/tex] will give =[tex]\frac{2}{2}\times 0.01=0.01moles[/tex] of [tex]H_2O[/tex]
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 101.325 kPa = 1 atm
V = Volume of gas = ?
n = number of moles = 0.01
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]40^0C=(40+273)K=313K[/tex]
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{0.01mol\times 0.0821L atm/K mol\times 313K}{1atm}=0.26L[/tex]
Thus the volume of product formed is 0.26 L
