what is the molarity of ZnCl2 that forms when 30.0 g of zinc completely reacts with CuCl2 according to the following reaction? Assume a final volume of 285 mL.
Zn(s) + CuCl2(aq)--> ZnCl2(aq) + Cu(s)
Use the following example of a stoichiometry problem to convert 30.0 g Zn to moles ZnCl2.Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Then M = moles/L (0.285 L from the problem).
To find the molarity of ZnCl2, we need to use the given mass of zinc (Zn) and the final volume of the solution.
First, let's determine the number of moles of zinc (Zn) present in 30.0 g of zinc. We can do this by dividing the mass of zinc by its molar mass. The molar mass of zinc (Zn) is 65.38 g/mol.
Moles of Zn = Mass of Zn / Molar mass of Zn
Moles of Zn = 30.0 g / 65.38 g/mol
Next, we have to determine the stoichiometric ratio between Zn and ZnCl2 using the balanced chemical equation. From the equation, we can see that 1 mole of Zn corresponds to 1 mole of ZnCl2.
Now, we need to convert the final volume of the solution from milliliters (mL) to liters (L) to use it in the molarity calculation. There are 1000 mL in 1 L.
Volume of solution = 285 mL / 1000 mL/L
Finally, we can calculate the molarity of ZnCl2 by dividing the moles of ZnCl2 by the volume of the solution in liters.
Molarity (M) of ZnCl2 = Moles of ZnCl2 / Volume of solution
Now that we have explained the steps, let's calculate the final answer.
Moles of Zn = 30.0 g / 65.38 g/mol = 0.4586 mol
Volume of solution = 285 mL / 1000 mL/L = 0.285 L
Molarity (M) of ZnCl2 = 0.4586 mol / 0.285 L
Therefore, the molarity of ZnCl2 that forms is approximately 1.61 M.