Answer: 0.367 moles of Zn will be needed to form 50.0 grams of ZnCl2.
Explanation:
To determine the number of moles of Zn required to form 50.0 grams of ZnCl2, we need to first find the molecular weight of ZnCl2, which can be calculated by adding the atomic weights of zinc (Zn) and two chlorine (Cl) atoms:
Molecular weight of ZnCl2 = atomic weight of Zn + 2 x atomic weight of Cl
= 65.38 + 2 x 35.45
= 136.28 g/mol
The number of moles of ZnCl2 can then be calculated by dividing the given mass (50.0 g) by the molecular weight (136.28 g/mol):
Number of moles of ZnCl2 = mass of ZnCl2 / molecular weight of ZnCl2
= 50.0 g / 136.28 g/mol
= 0.367 mol
Zinc (Zn) has a molar ratio of 1:1 with ZnCl2. This means that for every one mole of ZnCl2, one mole of Zn is required. Therefore, the number of moles of Zn needed to form 0.367 mol of ZnCl2 is also 0.367 mol.