If you are using M for mols, that won't get it. M stands for molarity. I will assume you meant mols. You have four limiting reagent (LR) problems here. I'll do the first two for you; you follow the plan to do the other two.
Zn + 2HCl ==> ZnCl2 + H2
a.
1. mols HCl = M x L = 0.2 x 0.025 = 0.005
2. mols Zn = g/atomic mass = 0.3/65.4 = approx 0.0046. NOte that you need to redo ALL of these calculations since I've estimated here and there.
3. Now you need to convert mols HCl and mols Zn to mols ZnCl2. Use the coefficients in the balanced equation to do this.
HCl: 0.005 mols HCl x (1 mol ZnCl2/2 mols HCl) = 0.0025 mols ZnCl2 formed IF WE HAD all of the Zn needed.
Zn: 0.0046 x (1 mol ZnCl2/1 mol Zn) = 0.0046 IF WE HAD all of the HCl we needed.
Note that you have two answers for ZnCl2 mols. Which do you choose? The correct value in LR problems is ALWAYS the smaller number. So mols ZnCl2 formed will be 0.0025 mols.
For b part.
1. mols HCl = 0.5 x 0.025 = 0.0125
mols Zn still approx 0.046.
2. Convert mols HCl to mols ZnCl2 and convert mols Zn to mols ZnCl2.
0.0125 mols HCl x (1 mol ZnCl2/2 mols HCl) = 0.00625 mols ZnCl2 formed IF WE HAD all of the Zn we needed.
For Zn it is 0.0046 x (1 mol ZnCl2/1 mol Zn) = 0.0046 mols ZnCl2 IF WE HAD all of the HCl we needed.
Here the smaller number is 0.0046 and that is the # mols ZnCl2 formed.
c and d are done the same way.
Post your work if you get stuck.
When zinc metal is reacted with hydrochloric acid, zinc chloride and hydrogen gas are produced. If 25ml of 0.2M HCl and 0.3g of zinc metal is used: a. how many M of zinc chloride will be produced? b. how many M of zinc chloride will be produced if 0.5M of HCl was used? c. how many M of zinc chloride will be produced if 1M of HCl was used? d. how many M of zinc chloride will be produced if 2M of HCl was used?
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