What mass of Zn can be dissolved by 558 mL of 3.01M HCl?
If I did it correctly is should be 110 g Zn
how do you figure that out?
Here is an example of a stoichiometry problem solved. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
if you haven't figured it out yet, go from the mL to L then to moles and then grams.
558mL x 1 L/1000mL x 3.01mol/1 L x 65.39 g/1 mol
hope this helps!!!
To calculate the mass of Zn that can be dissolved by 558 mL of 3.01M HCl, we need to use the concept of stoichiometry and balanced chemical equations.
The balanced chemical equation for the reaction between Zn and HCl is:
Zn + 2HCl → ZnCl2 + H2
From this equation, we can see that 1 mole of Zn reacts with 2 moles of HCl, producing 1 mole of ZnCl2 and 1 mole of H2.
Now, let's calculate the moles of HCl in 558 mL of 3.01M HCl.
Moles of HCl = (volume of HCl in liters) x (molarity of HCl)
= 0.558 L x 3.01 mol/L
= 1.67958 moles
Since the stoichiometry of the reaction is 1:2 (Zn:HCl), the moles of Zn required to react with 1.67958 moles of HCl is 1/2 of that:
Moles of Zn = (moles of HCl) / 2
= 1.67958 / 2
= 0.83979 moles
The molar mass of Zn is 65.38 g/mol. To calculate the mass of Zn, multiply the moles of Zn by its molar mass:
Mass of Zn = (moles of Zn) x (molar mass of Zn)
= 0.83979 mol x 65.38 g/mol
= 54.94 g
Therefore, approximately 54.94 grams of Zn can be dissolved by 558 mL of 3.01M HCl.