Respuesta :
Answer: 0.147 grams
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Al=\frac{1.33g}{27g/mol}=0.049moles[/tex]
[tex]\text{Moles of} NaOH=\frac{4.25g}{40g/mol}=0.106moles[/tex]
[tex]2Al(s)+2NaOH(aq)+6H_2O(l)\rightarrow Na(Al(OH)_4(aq)+3H_2(g)[/tex]
According to stoichiometry :
2 moles of [tex]Al[/tex] require = 2 moles of [tex]NaOH/tex]
Thus 0.049 moles of [tex]Al[/tex] will require=[tex]\frac{2}{2}\times 0.049=0.049moles[/tex] of [tex]NaOH[/tex]
Thus [tex]Al[/tex] is the limiting reagent as it limits the formation of product and [tex]NaOH[/tex] is the excess reagent.
As 2 moles of [tex]Al[/tex] give = 3 moles of [tex]H_2[/tex]
Thus 0.049 moles of [tex]Al[/tex] give =[tex]\frac{3}{2}\times 0.049=0.0735moles[/tex] of [tex]H_2[/tex]
Mass of [tex]H_2=moles\times {\text {Molar mass}}=0.0735moles\times 2g/mol=0.147g[/tex]
Thus 0.147 g of [tex]H_2[/tex] will be produced from the given masses of both reactants.
