Answer: The volume of hydrogen produced is 0.122 L
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of solute}={\text{Molarity}}\times {\text{Volume in L}}[/tex]
[tex]\text{Moles of} Mg=\frac{5.00g}{24g/mol}=0.208moles[/tex]
[tex]\text{Moles of} HCl=[/tex][tex]0.100moldm^{-3}\times 0.1dm^3=0.01mol[/tex]
The balanced chemical equation is:
[tex]Mg+2HCl\rightarrow MgCl_2+H_2[/tex]
According to stoichiometry :
2 moles of require = 1 mole of
Thus 0.01 moles of [tex]HCl[/tex] will require=[tex]\frac{1}{2}\times 0.01=0.005moles[/tex] of [tex]Mg[/tex]
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 2 moles of give = 1 mole of
Thus 0.01 moles of [tex]HCl[/tex] give =[tex]\frac{1}{2}\times 0.01=0.005moles[/tex] of [tex]H_2[/tex]
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 1.0 atm (SATP)
V = Volume of gas = ?
n = number of moles = 0.005
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]25^0C=(25+273)K=298K[/tex] (SATP)
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{0.005mol\times 0.0820 L atm/K mol\times 298K}{1.0atm}=0.122L[/tex]
Thus volume of hydrogen produced is 0.122 L
