How many mole of H2 will be produced if 0.500 grams of magnesium is reacted with 10.00mL of 2.0 M HCl?
I've concluded the answer is 0.02 moles of H2, but I'm doubting myself a lot because I don't understand the aspects of the balanced chemical equation and how that affects the outcome of the answer. Here is my work, please tell me where I've gone wrong in my work, and so forth. Thanks.
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2HCl + Mg ---> H2 + MgCl2
.02 mol HCl + .02057 mol Mg --->
0.03mol MgCl2 +0.02mol H2 +0.01057mol Mg
Mg Conversion to moles
0.5g X 1mol Mg/24.305g = 0.02057 mol Mg
HCl Conversion to moles
2.0M HCl = n/0.010L = 0.02moles HCl
Stoichiometry ~
2HCl/1Mg = .02moles HCl/X
X = .01 moles Mg will be consumed
Left Over Mg
.02057 mole Mg - .01 mole Mg = .01057moles Mg after reaction
Moles of MgCl2
.01 moles Mg + .02 moles Cl = .03 moles MgCl2 <-- I think this is wrong and should become .01 moles of MgCl2 because of the 2 on Cl.
moles of H2
Since all 10mL of 2.0M HCl is used up, then .02 moles of H2 is left. <-- I feel this is wrong because of the 2 of Hydrogen, so it should be .01 moles, not .02
Please help me in my confusion
This is a limiting reagent problem; we know that because amounts are given for BOTH reactants.
Mg + 2HCl ==> MgCl2 + H2
If we had ONLY Mg and all of the HCl needed, then 0.02057 moles Mg x (1 mole H2/1 mole mg) = 0.02057 moles H2 formed.
If we had 0.02 moles HCl and all the Mg needed, then 0.02 moles HCl x (1 mole H2/2 moles HCl) = 0.02 x 1/2 = 0.01 moles H2 formed.
Both answers can't be right; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Therefore, Mg is in excess and all of the HCl is consumed.
moles
Dr,
I have another question, but first, thank you for showing me that I was on the right track. I didn't understand how the moles of HCl were distributed. I thought .02 moles of HCl meant that there were .02 moles of H and .02 moles of Cl. Now I know that HCl add up to .02 moles.
My second question is, how would you find the volume of H2 gas at 1.0atm and 22°C? I've never learned this in class, but must find it for lab. Could you send me in the right direction, or perhaps give me a formula to follow?
Do you know n = number of moles. Then use PV = nRT and substitute 1 atm for P, V = ?, n = whatever, R is 0.08206 and T is 22 C converted to kelvin.
OR you may be measuring the volume of gas evolved from a reaction (such as Mg + 2HCl). In that case, you measure volume at room P and T and correct it to standard conditions of 1 atm and 273 K. Usually the gas is collected over water so you use P as room P-vapor pressure H2O at T of the room.
Thank you Dr. The former is needed to find the volume of H2 in this case. I expect that once I perform the lab, the latter will be needed.
You've helped me a great deal. You may expect to see me more often. I'm in your care for this semester, amongst any other Chemistry tutors.
To solve this problem, we need to first determine the limiting reagent. The limiting reagent is the reactant that gets completely consumed and determines the amount of product formed.
In this case, the balanced chemical equation is:
2HCl + Mg ---> H2 + MgCl2
From the equation, we can see that for every 1 mole of Mg, 1 mole of H2 is produced. So, if we know the amount of Mg in moles, we can directly determine the amount of H2 produced.
Let's examine your calculations step by step:
1. Mg Conversion to moles:
0.5g X (1 mol Mg / 24.305g) = 0.02057 mol Mg
This step is correct.
2. HCl Conversion to moles:
2.0 M HCl = 2.0 mol HCl / 1 L
10.00 mL = 0.01000 L
0.020 mol HCl X 0.010 L = 0.00020 mol HCl
This step is also correct.
3. Stoichiometry:
You correctly identified that the stoichiometric ratio is 2:1 for HCl:Mg. However, the calculation for the moles of Mg consumed is incorrect. Since the stoichiometric ratio is 2:1, we need to divide the moles of HCl by 2 to find the moles of Mg consumed. The correct calculation should be:
0.00020 mol HCl / 2 = 0.00010 mol Mg consumed
4. Leftover Mg:
0.02057 mol Mg - 0.00010 mol Mg = 0.02047 mol Mg
This step is correct.
5. Moles of MgCl2:
The stoichiometric ratio between Mg and MgCl2 is 1:1, meaning that for every mole of Mg consumed, one mole of MgCl2 is formed. Therefore, the correct calculation is:
0.00010 mol Mg + 0.00010 mol Cl = 0.00010 mol MgCl2
You are correct that this should be 0.01 moles of MgCl2, not 0.03 moles.
6. Moles of H2:
Since the stoichiometric ratio between Mg and H2 is 1:1, the number of moles of H2 produced is equal to the number of moles of Mg consumed. Therefore, the correct answer is:
0.00010 mol H2
So, in conclusion, the correct answer is that 0.00010 moles (or 0.02 moles) of H2 will be produced when 0.500 grams of magnesium reacts with 10.00 mL of 2.0 M HCl.