What is the molarity of ZnCl2 that forms when 30.0g 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)
Please Explain!! Thanks!!
To find the molarity of ZnCl2, we first need to calculate the number of moles of Zn reacted with CuCl2. We can use the molar mass of Zn to convert the given mass of Zn to moles.
1. Calculate the molar mass of Zn:
The molar mass of Zn is the sum of the atomic masses of its elements, which is 65.38 g/mol (from the periodic table).
2. Convert the mass of Zn to moles:
Moles of Zn = mass of Zn / molar mass of Zn
Moles of Zn = 30.0 g / 65.38 g/mol = 0.458 mol
Now, we can use the stoichiometry of the balanced equation to determine the number of moles of ZnCl2 produced.
According to the balanced equation, the mole ratio between Zn and ZnCl2 is 1:1. This means that for every 1 mole of Zn, we get 1 mole of ZnCl2.
3. Calculate the moles of ZnCl2:
Moles of ZnCl2 = Moles of Zn = 0.458 mol
Finally, we can calculate the molarity of ZnCl2 using the definition of molarity:
Molarity = Moles of solute / Volume of solution (in liters)
4. Convert the given volume from mL to liters:
Volume of solution = 285 mL = 285/1000 L = 0.285 L
5. Calculate the molarity of ZnCl2:
Molarity of ZnCl2 = Moles of ZnCl2 / Volume of solution
Molarity of ZnCl2 = 0.458 mol / 0.285 L ≈ 1.61 M
Therefore, the molarity of ZnCl2 formed when 30.0g of zinc completely reacts with CuCl2 in a final volume of 285 mL is approximately 1.61 M.