Refer to 3CuCl2 (aq)+2Na3PO4 (aq) = Cu3(PO4)2+6NaCl (aq)
If a student started with 18 grams of CuCl2 and enough Na3PO4 to completely react, how many grams of Cu3(PO4)2 will (theoretically) be produced?
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the number of grams of Cu3(PO4)2 produced in the reaction, we need to calculate the theoretical yield. The theoretical yield refers to the maximum amount of product that can be obtained under ideal conditions.
We can start by calculating the molar masses of the compounds involved in the reaction:
- CuCl2: 63.55 g/mol (molar mass of copper: 63.55 g/mol) + (2 * 35.45 g/mol) (molar mass of chlorine: 35.45 g/mol)
- Na3PO4: (3 * 22.99 g/mol) (molar mass of sodium: 22.99 g/mol) + (1 * 30.97 g/mol) (molar mass of phosphorus: 30.97 g/mol) + (4 * 16.00 g/mol) (molar mass of oxygen: 16.00 g/mol)
Next, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the amount of product that can be formed. In this case, we will compare the moles of CuCl2 and Na3PO4 to see which one is limiting.
1. Calculate the moles of CuCl2:
Moles = Mass / Molar mass
Moles of CuCl2 = 18 g / (63.55 g/mol + 2 * 35.45 g/mol)
2. Calculate the moles of Na3PO4:
Moles of Na3PO4 = ? (since we don't have the mass)
Since the balanced equation tells us that the stoichiometric ratio between CuCl2 and Na3PO4 is 3:2, we can calculate the moles of Na3PO4 needed to react with the moles of CuCl2 calculated in step 1.
Moles of Na3PO4 = (Moles of CuCl2) * (2 / 3)
Now, assuming there is enough Na3PO4 to completely react with CuCl2, we can calculate the moles of Cu3(PO4)2 that will be produced.
Since the balanced equation tells us that the stoichiometric ratio between CuCl2 and Cu3(PO4)2 is 3:1, we can calculate the moles of Cu3(PO4)2 formed by multiplying the moles of CuCl2 by (1 / 3).
Finally, multiply the moles of Cu3(PO4)2 by its molar mass to find the theoretical yield (in grams).
I'm sorry, but without knowing the mass of Na3PO4, we cannot determine the number of grams of Cu3(PO4)2 produced in the reaction.