how many grams of Zn metal would you use to prepare 6.8g of zinc chloride according the reaction: Zn(s)+2HCl(aq)-> ZnCl2(aq)+H2(g)
To determine the number of grams of Zn metal required to prepare 6.8g of zinc chloride (ZnCl2), you can use the stoichiometry of the balanced chemical equation.
Step 1: Determine the molar mass of ZnCl2
ZnCl2 consists of one zinc atom (Zn) and two chlorine atoms (Cl). The atomic masses of Zn and Cl are approximately 65.4 g/mol and 35.5 g/mol, respectively.
Molar mass of ZnCl2 = (1 * molar mass of Zn) + (2 * molar mass of Cl)
= (1 * 65.4) + (2 * 35.5)
= 65.4 + 71.0
= 136.4 g/mol
Step 2: Calculate the number of moles of ZnCl2
The number of moles (n) of ZnCl2 can be determined using the formula:
n = mass / molar mass
n = 6.8 g / 136.4 g/mol
n ≈ 0.0498 mol
Step 3: Use the stoichiometry of the reaction to find the number of moles of Zn metal
According to the balanced chemical equation, the stoichiometric ratio between Zn and ZnCl2 is 1:1. This means that 1 mole of Zn reacts with 1 mole of ZnCl2.
Therefore, the number of moles of Zn required is also approximately 0.0498 mol.
Step 4: Convert moles of Zn to grams
Using the molar mass of Zn (65.4 g/mol), multiply the number of moles of Zn by its molar mass to get the mass in grams:
mass = moles * molar mass
= 0.0498 mol * 65.4 g/mol
≈ 3.26 g
Thus, you would need approximately 3.26 grams of zinc (Zn) metal to prepare 6.8 grams of zinc chloride (ZnCl2) according to the given reaction.