Respuesta :
72.0gAl× 2 mol Al × 2 mol alcl3 × (grams alcl3)
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(grams Al) 2 mol Al 2 mol alcl3
same goes for HCL, remember to use mole to mole ratio
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(grams Al) 2 mol Al 2 mol alcl3
same goes for HCL, remember to use mole to mole ratio
Answer : The mass of [tex]AlCl_3[/tex] produced will be, 46.935 grams.
Explanation : Given,
Mass of [tex]Al[/tex] = 9.5 g
Mass of [tex]HCl[/tex] = 130 g
Molar mass of [tex]Al[/tex] = 26.98 g/mole
Molar mass of [tex]HCl[/tex] = 36.5 g/mole
Molar mass of [tex]AlCl_3[/tex] = 133.34 g/mole
First we have to calculate the moles of [tex]Al[/tex] and [tex]HCl[/tex].
[tex]\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=\frac{9.5g}{26.98g/mole}=0.352moles[/tex]
[tex]\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}=\frac{130g}{36.5g/mole}=3.56moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]2Al+6HCl\rightarrow 2AlCl_3+3H_2[/tex]
From the balanced reaction we conclude that
As, 2 moles of [tex]Al[/tex] react with 6 mole of [tex]HCl[/tex]
So, 0.352 moles of [tex]Al[/tex] react with [tex]\frac{6}{2}\times 0.350=0.704[/tex] moles of [tex]HCl[/tex]
From this we conclude that, [tex]HCl[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Al[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]AlCl_3[/tex].
As, 2 moles of [tex]Al[/tex] react to give 2 moles of [tex]AlCl_3[/tex]
So, 0.352 moles of [tex]Al[/tex] react to give 0.352 moles of [tex]AlCl_3[/tex]
Now we have to calculate the mass of [tex]AlCl_3[/tex].
[tex]\text{Mass of }AlCl_3=\text{Moles of }AlCl_3\times \text{Molar mass of }AlCl_3[/tex]
[tex]\text{Mass of }AlCl_3=(0.352mole)\times (133.34g/mole)=46.935g[/tex]
Therefore, the mass of [tex]AlCl_3[/tex] produced will be, 46.935 grams.
