Respuesta :
Answer:
Part A: 5.899x10^-3 moles of Al
Part B: 1.573 g of AlBr3
Explanation:
Part A: We have to obtain the volume of the piece of aluminium; all sides of the square must be in cm. Then, use the density to obtain the mass.
[tex]0.590 mm (\frac{1 cm}{10 mm}) = 0.059 cm[/tex]
[tex]V= lxlxl= (1cm)(1cm)(0.059 cm)= 0.059 cm^3.[/tex]
0.059 is the volume of the Al udes for the reaction. Now, to oabtain the moles:
[tex]0.059 cm^3 Al(\frac{2.699 g Al}{1 cm^3 Al})(\frac{1 mol Al}{27 g Al})= 5.899x10^-3 moles[/tex]
Part B: To obatin the mass of AlBr3, we need the balanced chemical equation:
2Al + 3Br2 → 2AlBr3
We assume bromine (Br2) is in excess, therefore, we calculate the aluminum bromide formed from the Al:
[tex]5.898 x 10^-3 moles Al (\frac{2 moles AlBr3}{2 moles Al}) (\frac{266.7 g AlBr3}{1 mol AlBr3}) = 1.573 g[/tex] of Al
Part A: [tex]5.899*10^{-3}[/tex] moles of Al
Part B: [tex]1.573 g[/tex] of AlBr₃
Part A:
We have to obtain the volume of the piece of aluminum; all sides of the square must be in cm. Then, use the density to obtain the mass.
1mm = 10 cm
[tex]0.590mm*\frac{1cm}{10mm}=0.059cm[/tex]
As we know, Volume is the product of length, breadth and height, It can be written as:
[tex]\text{Volume}=l*b*h\\\\\text{Volume}=1*1*0.059\\\\\text{Volume}=0.59cm^3[/tex]
Calculation for moles:
0.059 is the volume of the Al is used for the reaction. Now, to obtain the moles:
[tex]0.509cm^3Al*\frac{2.699g Al}{1cm^3Al} *\frac{1mol Al}{27g Al}=5.899*10^{-3}moles[/tex]
Part B:
To obtain the mass of AlBr₃, we need the balanced chemical equation:
2Al + 3Br₂ → 2AlBr₃
We assume bromine (Br₂) is in excess, therefore, we calculate the aluminum bromide formed from the Al:
[tex]5.89*10^{-3}\text{ moles of Al }*\frac{\text{ 2 moles of } AgBr_3}{\text{ 2 moles of Al} }*\frac{266.7g AlBr_3}{1 mol AlBr_3} =1.573 \text{ g of Al }[/tex]
Thus,
Part A: [tex]5.899*10^{-3}[/tex] moles of Al
Part B: [tex]1.573 g[/tex] of AlBr₃
Find more information about calculation of moles here:
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