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2.What is the limiting reagent if 0.5 g Al is reacted with 3.5 g CuCl2? (Reminder: CuCl2 is a dihydrate)

2.What is the limiting reagent if 0.5 g Al is reacted with 3.5 g CuCl2? (Reminder-example-1
User Mateusmaso
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1 Answer

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3) In question 1, we find the following balanced equation:

2 Al + 3 CuCl2.2H2O ---> 3 Cu + 2 AlCl3 + 6 H2O

Step 1 - Let's find out how many moles of Al and CuCl2 reacts

For this, we will use the following formula: mole = mass/molar mass

Molar mass of Al: 26.981539 g/mol

Molar mass of CuCl2.2H2O: 147.01 g/mol

Al:

mass = 0.5 g

molar mass = 26.981539 g/mol

mole = 0.5/26.981539

mole = 0.018531189

CuCl2.2H2O:

mass: 3.5 g

Molar mass: 147.01 g/mol

mole = 0.023807904

Step 2 - Let's see which one is the limiting reactant.

2 moles Al --- 3 moles CuCl2.2H2O

0.018531189 moles Al --- x

x = 0.02779678

2 moles Al --- 3 moles CuCl2.2H2O

x moles Al --- 0.023807904 moles CuCl2.2H2O

x = 0.015871936

CuCl2.2H2O is the limiting reactant. It means that 0.023807904 moles CuCl2.2H2O react with 0.015871936 moles of Al and produce 0.023807904 moles of Cu.

Step 3 - Let's transform 0.023807904 moles of Cu into grams. For this, we use the following formula: mass = mole x molar mass

molar mass of Cu = 63.546

mole = 0.023807904 moles

mass = 0.023807904 x 63.546

mass = 1.5129 g

Answer: The theoretical yield of copper produced is 1.5129 g

User Gavin Terrill
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