How many grams of aluminum would react completely with 17.50 grams of copper (II) chloride according to the following equation?
Al(s) + CuCl2(aq) ---> AlCl3(aq) + Cu(s
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To determine the number of grams of aluminum that would react completely with 17.50 grams of copper (II) chloride, you need to use stoichiometry and the balanced chemical equation.
Step 1: Write and balance the chemical equation:
2 Al(s) + 3 CuCl2(aq) ---> 2 AlCl3(aq) + 3 Cu(s)
Step 2: Find the molar mass of CuCl2:
Copper (Cu) has a molar mass of 63.55 g/mol, and chlorine (Cl) has a molar mass of 35.45 g/mol. Since there are two chlorine atoms in copper (II) chloride, the molar mass of CuCl2 is:
(2 * 35.45 g/mol) + 63.55 g/mol = 134.45 g/mol
Step 3: Convert the mass of CuCl2 to moles:
The given mass of CuCl2 is 17.50 grams. To convert grams to moles, divide the mass by the molar mass:
17.50 g / 134.45 g/mol ≈ 0.1301 mol
Step 4: Use stoichiometry to determine the moles of aluminum:
According to the balanced equation, 3 moles of CuCl2 react with 2 moles of Al. Therefore, the moles of Al can be calculated using the mole ratio:
0.1301 mol CuCl2 * (2 mol Al / 3 mol CuCl2) ≈ 0.0867 mol Al
Step 5: Convert moles of Al to grams:
To convert moles to grams, multiply the moles by the molar mass of aluminum, which is 26.98 g/mol:
0.0867 mol Al * 26.98 g/mol ≈ 2.34 grams
Therefore, approximately 2.34 grams of aluminum would react completely with 17.50 grams of copper (II) chloride according to the given balanced equation.