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3.69 A piece of Aluminum foil 1.00cm^2 and 0.550mm thick is allowed react with bromine to form aluminum bromide. (a) How many moles of aluminum were used? (D=2.699g/cm^3) (b) How many grams of aluminum bromide form , assuming reacts completely? (Hint: Write the balanced equation first)

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3.69 A piece of Aluminum foil 1.00cm^2 and 0.550mm thick is allowed react with bromine to form aluminum bromide. (a) How many moles of aluminum were used? (D=2.699g/cm^3) (b) How many grams of aluminum bromide form , assuming reacts completely? (Hint: Write the balanced equation first)

3.69 A piece of Aluminum foil 1.00cm^2 and 0.550mm thick is allowed react with bromine to form aluminum bromide. (a) How many moles of aluminum were used? (D=2.699g/cm^3) (b) How many grams of aluminum bromide form , assuming reacts completely? (Hint: Write the balanced equation first)

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MiltonElite · Tutor for 8 years

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(a) <br />1. Calculate the volume of aluminum foil:<br />\[ \text{Volume} = \text{Area} \times \text{Thickness} = 1.00 \, \text{cm}^2 \times 0.0550 \, \text{cm} = 0.0550 \, \text{cm}^3 \]<br /><br />2. Calculate the mass of aluminum using density:<br />\[ \text{Mass} = \text{Volume} \times \text{Density} = 0.0550 \, \text{cm}^3 \times 2.699 \, \text{g/cm}^3 = 0.148445 \, \text{g} \]<br /><br />3. Convert mass to moles (molar mass of Al = 26.98 g/mol):<br />\[ \text{Moles of Al} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.148445 \, \text{g}}{26.98 \, \text{g/mol}} = 0.0055 \, \text{mol} \]<br /><br />(b)<br />1. Write the balanced chemical equation:<br />\[ 2 \text{Al} + 3 \text{Br}_2 \rightarrow 2 \text{AlBr}_3 \]<br /><br />2. Calculate moles of aluminum bromide formed (1:1 ratio with aluminum):<br />\[ \text{Moles of AlBr}_3 = 0.0055 \, \text{mol} \]<br /><br />3. Convert moles of aluminum bromide to grams (molar mass of AlBr\(_3\) = 266.69 g/mol):<br />\[ \text{Mass of AlBr}_3 = \text{Moles} \times \text{Molar Mass} = 0.0055 \, \text{mol} \times 266.69 \, \text{g/mol} = 1.4668 \, \text{g} \]<br /><br /># Answer:<br />(a) 0.0055 mol of aluminum<br /><br />(b) 1.4668 g of aluminum bromide

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To solve this problem, we need to determine the moles of aluminum used and the grams of aluminum bromide formed. We start by calculating the volume and mass of aluminum, then use stoichiometry based on the balanced chemical equation for the reaction between aluminum and bromine.
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