Respuesta :
Answer: 3.02 L of [tex]H_2[/tex] will be produced from the given masses of both reactants.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Zn=\frac{10g}{65g/mol}=0.15moles[/tex]
[tex]\text{Moles of} HCl=\frac{10g}{36.5g/mol}=0.27moles[/tex]
[tex]Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)[/tex]
According to stoichiometry :
2 moles of [tex]HCl[/tex] require = 1 mole of [tex]Zn[/tex]
Thus 0.27 moles of [tex]HCl[/tex] will require=[tex]\frac{1}{2}\times 0.27=0.135moles[/tex] of [tex]Zn[/tex]
Thus [tex]HCl[/tex] is the limiting reagent as it limits the formation of product and [tex]Zn[/tex] is the excess reagent.
As 2 moles of [tex]HCl[/tex] give = 1 moles of [tex]H_2[/tex]
Thus 0.27 moles of [tex]HCl[/tex] give =[tex]\frac{1}{2}\times 0.27=0.135moles[/tex] of [tex]H_2[/tex]
Volume of [tex]H_2=moles\times {\text {Molar Volume}}=0.135moles\times 22.4g/L=3.02L[/tex]
Thus 3.02 L of [tex]H_2[/tex] will be produced from the given masses of both reactants.
