How many grams of aluminum chloride are produced when 18 g of aluminum are reacted with an excess of hydrochloric acid?
2AL(s) + 6HCl(aq) ---> 2AlCl3(aq) + 3H2(g)
mols Al = grams/molar mass
Convert mols Al to mols AlCl3 using the coefficients in the balanced equation.
Convert mols AlCl3 to grams. g = mols x molar mass.
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To find out how many grams of aluminum chloride are produced, we need to use the stoichiometry of the balanced chemical equation.
Based on the balanced equation:
2 moles of aluminum react with 6 moles of hydrochloric acid to produce 2 moles of aluminum chloride.
Step 1: Convert the mass of aluminum to moles.
We need to convert the 18 grams of aluminum to moles. The molar mass of aluminum (Al) is 26.98 g/mol.
Number of moles of aluminum = Mass of aluminum / molar mass of aluminum
Number of moles of aluminum = 18 g / 26.98 g/mol
Number of moles of aluminum = 0.668 mol (rounded to 3 decimal places)
Step 2: Determine the moles of aluminum chloride produced.
Since the stoichiometric ratio between aluminum and aluminum chloride is 2:2, we can conclude that the number of moles of aluminum chloride produced is also 0.668 moles.
Step 3: Convert the moles of aluminum chloride to grams.
The molar mass of aluminum chloride (AlCl3) is 133.34 g/mol.
Mass of aluminum chloride = Number of moles of AlCl3 x molar mass of AlCl3
Mass of aluminum chloride = 0.668 mol x 133.34 g/mol
Mass of aluminum chloride = 89.02 g (rounded to 2 decimal places)
Therefore, 18 grams of aluminum will produce 89.02 grams of aluminum chloride when reacted with an excess of hydrochloric acid.
To find the number of grams of aluminum chloride produced, we need to use the given balanced equation and the mole-to-mass ratio. Here's how you can calculate it:
Step 1: Find the molar mass of aluminum (Al).
The atomic mass of aluminum (Al) is approximately 27 g/mol.
Step 2: Set up a ratio using the balanced equation.
From the balanced equation, we see that 2 moles of aluminum react to produce 2 moles of aluminum chloride. This means there is a 1:1 mole ratio between aluminum and aluminum chloride.
Step 3: Convert grams of aluminum to moles.
To convert grams of aluminum to moles, divide the given mass of aluminum by its molar mass:
18 g Al * (1 mol Al / 27 g Al) = 0.67 mol Al
Step 4: Use the mole ratio to find the moles of aluminum chloride.
Since the mole ratio between aluminum and aluminum chloride is 1:1, the moles of aluminum chloride will be the same as the moles of aluminum.
Moles of AlCl3 = 0.67 mol Al
Step 5: Convert moles of aluminum chloride to grams.
To convert moles of aluminum chloride to grams, multiply the moles by its molar mass. The molar mass of aluminum chloride (AlCl3) is approximately 133.34 g/mol.
Grams of AlCl3 = 0.67 mol AlCl3 * (133.34 g AlCl3 / 1 mol AlCl3) = 89.31 g AlCl3
Therefore, approximately 89.31 grams of aluminum chloride are produced when 18 grams of aluminum are reacted with an excess of hydrochloric acid.