if unlimited iron is available to react with 34.0g of copper (II) chloride, how many g of copper metal would be produced?
Here is a worked example, Follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
how do you know what the actual yield will be?!
The example gives theoretical yield and that's all you can calculate from the data you posted. If you have the percent yield you can determine actual yield by
actual yield = theor yield x (%yield/100) = ??; otherwise, it can be done only by performing the experiment and measuring the actual yield.
To determine the mass of copper metal produced when unlimited iron reacts with 34.0g of copper (II) chloride, we need to use stoichiometry, which relates the amounts of reactants and products in a chemical reaction.
First, we write the balanced chemical equation for the reaction between iron and copper (II) chloride:
Fe + CuCl2 → FeCl2 + Cu
From the balanced equation, we can see that 1 mole of iron reacts with 1 mole of copper (II) chloride to produce 1 mole of copper metal.
To find the molar mass of copper (II) chloride (CuCl2), we add the atomic masses of its constituent elements:
Atomic mass of Cu = 63.55 g/mol
Atomic mass of Cl = 35.45 g/mol (there are two chlorine atoms in CuCl2)
Molar mass of CuCl2 = 63.55 g/mol + 2 * 35.45 g/mol = 134.45 g/mol
Next, we calculate the number of moles of copper (II) chloride using its mass:
Number of moles = Mass / Molar mass
Number of moles = 34.0 g / 134.45 g/mol ≈ 0.253 mol
Since the reaction is stoichiometric, 0.253 moles of copper (II) chloride will react to produce the same number of moles of copper metal.
Finally, we calculate the mass of the copper metal produced using its molar mass:
Mass = Number of moles x Molar mass
Mass = 0.253 mol x 63.55 g/mol ≈ 16.1 g
Therefore, approximately 16.1 grams of copper metal would be produced when unlimited iron reacts with 34.0 grams of copper (II) chloride.