How plentiful grams of zinc metal will take action beside 24.5 mL of 2.0 M hydrochloric sour, HCl, forming zinc chloride?
Answers:
First, lets set up the equation for the synthesis reaction:
Zn+HCl ----> ZnCl2 + H2
NOTE: H exists as a diatomic molecule surrounded by nature. Also, be sure that zinc is above H in the stir series (it is).
Now lets balance; It looks similar to everything is balanced except for the HCl on the left and the H2 on the right, so we add on a 2 in front of HCl on the left:
Zn+2HCl ----> ZnCl2 + H2
Lets check that we hold equal numbers of each atom on each side:
Left side: 1 Zn, 2 H, 2 Cl
Right side: 1 Zn, 2 Cl, 2 H
The equation is indeed suspended.
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Now, using the molarity and volume of the HCl, we can figure out how many mols of HCl are used:
Molarity=mols/liter
2=mols/.0245 (24.5mL=.0245L)
.049 mols HCl
Now let look at our equation again:
Zn+2HCl ----> ZnCl2 + H2
The mol ratio of Zn to HCl is 1:2
.049 mols HCl (1 mol Zn / 2 mols HCl)
The mols HCl unit cancels out, disappearing .0245 mols Zn
Now we can do the last step; finding how many grams of Zn be used. The molar mass of Zn is 65.39.
.0245 mols Zn (65.39g / 1 mol Zn)
The mols Zn unit cancels out, departure 1.602 (rounded) g Zn.
Approximately 1.6 grams of Zn will react with the HCl.
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If you call for further explanation or any other help, feel free to ask.
Moles HCl = 0.0245 L x 2.0 M= 0.049
Zn + 2 HCl = ZnCl2 + H2
the ratio between Zn and HCl is 1 : 2
moles Zn needed = 0.049/2=0.0245
mass Zn = 0.0245 mol x 65.39 g/mol=1.6
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