if 25.o g of Al are reacted with an excess of amount of Cl, what is the maximum amount of aluminum chloride that can be produced using the following reaction. 2 Al + 3 Cl2 - 2 AlCl3
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To find the maximum amount of aluminum chloride that can be produced, you need to determine which reactant limits the reaction. In this case, we have an excess amount of chlorine (Cl2), which means the limiting reactant is aluminum (Al).
To calculate the moles of aluminum (Al), you can use the formula:
moles = mass / molar mass = 25.0 g / 26.98 g/mol = 0.926 mol
From the balanced equation, we know that 2 moles of aluminum react with 2 moles of aluminum chloride (AlCl3). Therefore, the moles of aluminum chloride produced will also be 0.926 mol.
Finally, to calculate the mass of aluminum chloride (AlCl3), you can use the formula:
mass = moles × molar mass = 0.926 mol × (26.98 g/mol + 3 × 35.45 g/mol) = 0.926 mol × 133.35 g/mol = 122.99 g
Therefore, the maximum amount of aluminum chloride that can be produced is 122.99 grams.