Answer: 0.771 g of [tex]H_2[/tex] will be produced
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe=\frac{14.4g}{56g/mol}=0.257moles[/tex]
The balanced chemical reaction is:
[tex]2Fe(s)+6HCl(aq)\rightarrow 2FeCl_3(aq)+3H_2(g)[/tex]
According to stoichiometry :
2 moles of [tex]Fe[/tex] produce = 3 moles of [tex]H_2[/tex]
Thus 0.257 moles of [tex]Fe[/tex] will produce=[tex]\frac{3}{2}\times 0.257=0.385moles[/tex] of [tex]H_2[/tex]
Mass of [tex]H_2=moles\times {\text {Molar mass}}=0.385moles\times 2g/mol=0.771g[/tex]
Thus 0.771 g of [tex]H_2[/tex] will be produced
