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 mass of zinc metal required to prepare a given amount of zinc chloride, we need to use stoichiometry. Stoichiometry allows us to balance the equation and relate the amounts of reactants and products.
Step 1: Balance the Chemical Equation
The equation is already balanced.
Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)
Step 2: Determine the Molar Masses
The molar mass of Zn is 65.38 g/mol, and the molar mass of ZnCl2 is 136.29 g/mol.
Step 3: Convert the Given Mass to Moles
Divide the given mass of zinc chloride (6.8 g) by its molar mass (136.29 g/mol) to convert it to moles:
6.8 g / 136.29 g/mol = 0.0499 mol
Step 4: Apply Stoichiometry
From the balanced equation, we can see that 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 amount of Zn needed to produce 0.0499 mol of ZnCl2 is also 0.0499 mol.
Step 5: Convert Moles of Zn to Grams
Multiply the number of moles by the molar mass of Zn to convert it to grams:
0.0499 mol × 65.38 g/mol = 3.28 g
So, you would need approximately 3.28 grams of Zn metal to prepare 6.8 grams of zinc chloride according to the given reaction.