- #1
Batmaniac
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Homework Statement
5.00g alloy of Magnesium and Aluminum is treated with excess HCl, forming MgCl2 and AlCl3 and 6.65L of H2 at 25 degrees celcius and 99.2kPa. What is the mass percent of Mg in the alloy?
Homework Equations
PV = nRT
The Attempt at a Solution
Using PV = nRT, I found the moles of H2 gas to be approximately 0.2663. Now I attempted to setup my overall reaction by combining the half reactions.
Mg + 2HCL --> MgCl2 + H2 (1)
2Al + 6HCL --> 2AlCl3 + 3H2 (2)
Adding them we obtain:
Mg + 2Al + 8HCL ---> MgCl2 + 2AlCl3 + 4H2
Using the above equation, I know that 1 mol of Mg reacts with 4 moles of H2, therefore x moles of Mg reacts with 0.2663 moles of H2. Therefore, the moles of Mg is approximately equal to .2663/4 = 0.066575 mol.
Multiplying the moles of Mg by it's molar mass, 24.31 g/mol gives you about 1.6g of Mg, which in turn if 32% of the 5g compound.
The problem is that the answer to the question isn't 32%, it's 16%. If it's 16% then my mole ratio was likely wrong because if you have a ratio of 1 mol of Mg for every 8 moles of H2, you get 16% as your answer, I just don't see how that ratio would come to be.
Furthermore, the charges are not balanced in my equations, is that a problem?
- Thanks