A sample of impure magnesium was analyzed by allowing it to react with excess HCl solution:
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
After 1.27 g of the impure metal was treated with 0.100 L of 0.768 M HCl, 0.0125 mol HCl remained. Assuming the impurities do not react, what is the mass % of Mg in the sample?
I converted the 1.27gMg to 0.0522moles Mg. Im just not sure where to go from there. please help me in the right direction.
Thanks!
Nope, you can't directly convert the 1.27 g sample to moles of Mg because it is impure or not 100% Mg.
Since there is some HCl left after the reaction, the impure Mg is the limiting reactant and HCl is in excess.
Note that the Mg used was impure.
To get the number of moles, we use its the relationship with Molarity:
M = n/V
where n = moles and V = liters of solution
We get the moles of HCl:
n, HCl = 0.768 * 0.100 = 0.0768 moles HCl
Since not all of this reacted (as we said above, HCl is in excess), we subtract this from the amount of HCl left:
0.0768 - 0.0125 = 0.0643 moles HCl (amount only reacted with Mg)
Using the stoich ratios from the reaction, for every 2 mol HCl reacted, 1 mol Mg reacted. So,
0.0643 moles HCl * (1 mol Mg / 2 mol HCl) = 0.03215 mol Mg (this is pure)
The molar mass of Mg is 24.3 g/mol. So the mass of pure Mg is therefore,
0.03215 mol * 24.3 g/mol = 0.781245 g pure Mg
Finally, we can get the % pure Mg in the sample:
0.781245 / 1.27 * 100 = 61.5 %
Hope this helps~ :3
You erred in calculating mols Mg.
What you have written is true only if the Mg metal were pure; however, it isn't.
mols HCl begin = M x L = 0.768 x 0.1 = ?
mols HCl remain = 0.0125
the difference is mols HCl used.
Then mols Mg is 1/2 that of HCl.
g Mg is mols x atomic mass Mg
%Mg = (g Mg/mass sample)*100 = ?
I get something like 61% but that isn't exact.
Well, it seems like you're almost there! You've determined that you have 0.0522 moles of Mg. Now, you have to find the moles of HCl that reacted with the Mg.
From the balanced equation, we can see that for every 1 mole of Mg, 2 moles of HCl react. So, the moles of HCl that reacted can be calculated by multiplying the moles of Mg by the stoichiometric ratio:
0.0522 moles Mg × (2 moles HCl / 1 mole Mg) = 0.1044 moles HCl
Now, we have to find the moles of HCl that remained. We're given that 0.100 L of 0.768 M HCl was used, so we can calculate the moles of HCl from the concentration and volume:
0.768 moles HCl per 1 L × 0.100 L = 0.0768 moles HCl
Now, we subtract the moles of HCl that reacted from the moles of HCl used initially to find the moles of HCl that remained:
0.0768 moles HCl - 0.1044 moles HCl = -0.0276 moles HCl
Since we can't have negative moles, it seems like we made an error. Let's try troubleshooting.
To find the mass % of Mg in the sample, you need to calculate the moles of MgCl2 formed from the reaction and then use that to determine the moles and mass of Mg.
1. Calculate the moles of HCl reacted:
Since 0.100 L of 0.768 M HCl was used, the moles of HCl can be calculated using the equation:
moles of HCl = volume (L) × concentration (mol/L)
moles of HCl = 0.100 L × 0.768 mol/L = 0.0768 mol HCl
2. Calculate the moles of HCl that remained:
Given that 0.0125 mol of HCl remained, subtract that from the initial moles of HCl:
moles of HCl reacted = initial moles of HCl - moles of HCl remaining
moles of HCl reacted = 0.0768 mol - 0.0125 mol = 0.0643 mol
3. Use the stoichiometry of the balanced equation to calculate the moles of MgCl2 formed:
From the balanced equation, the molar ratio between Mg and HCl is 1:2. This means that for every 1 mole of Mg, 2 moles of HCl are consumed.
Therefore, the moles of MgCl2 formed = 0.0643 mol HCl / 2 = 0.0322 mol MgCl2.
4. Calculate the moles of Mg:
Since the molar ratio between Mg and MgCl2 is 1:1, the moles of Mg will be the same as the moles of MgCl2:
moles of Mg = 0.0322 mol MgCl2.
5. Finally, calculate the mass % of Mg in the sample:
mass % of Mg = (moles of Mg / total moles of sample) × 100
mass % of Mg = (0.0322 mol / 0.0522 mol) × 100 = 61.72%
Therefore, the mass % of Mg in the impure sample is approximately 61.72%.
To find the mass % of Mg in the sample, you need to determine the mass of Mg present in the sample and then calculate the percentage.
First, let's calculate the number of moles of Mg reacted in the given equation:
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
From the equation, we can see that the ratio between Mg and HCl is 1:2. Therefore, the number of moles of HCl reacted is twice the number of moles of Mg reacted.
The initial number of moles of HCl in the solution can be calculated using its concentration and volume:
moles of HCl = concentration × volume
moles of HCl = 0.768 M × 0.100 L
moles of HCl = 0.0768 mol
Since 0.0125 mol of HCl remained, the number of moles of HCl reacted is:
moles of HCl reacted = initial moles of HCl - remaining moles of HCl
moles of HCl reacted = 0.0768 mol - 0.0125 mol
moles of HCl reacted = 0.0643 mol
Since the ratio of Mg to HCl is 1:2, the number of moles of Mg reacted is half the number of moles of HCl reacted:
moles of Mg reacted = 0.0643 mol / 2
moles of Mg reacted = 0.0322 mol
Now, we can calculate the mass of Mg in the sample using its molar mass:
mass of Mg = moles of Mg × molar mass of Mg
mass of Mg = 0.0322 mol × 24.31 g/mol (molar mass of Mg)
mass of Mg = 0.783 g
Finally, we can calculate the mass % of Mg in the sample:
mass % of Mg = (mass of Mg / mass of sample) × 100
mass % of Mg = (0.783 g / 1.27 g) × 100
mass % of Mg = 61.57%
Therefore, the mass % of Mg in the sample is approximately 61.57%.