What is the Limiting Reactant in this HCl and Mg Reaction?

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To determine the limiting reactant in the reaction between HCl and Mg, first calculate the moles of HCl in 25 mL of a 1M solution, which equals 0.025 moles. Using the reaction equation, 2 moles of HCl are required for every mole of Mg, establishing a 2:1 ratio. Next, convert the weights of Mg (0.15g, 0.3g, and 0.6g) to moles, using the molar mass of Mg (24.31 g/mol). Compare the moles of HCl available to the moles of Mg to identify the limiting reactant for each sample. This analysis will also allow for the calculation of the hydrogen gas produced in each reaction.
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Homework Statement


Procedure
1. Add 25 mL of HCL solution to each flask.
2. Weigh out 0.15g, 0.3g, and 0.6g of Mg ribbon and place each sample into its own balloon




Homework Equations



Mg + 2HCl --> H2 + MgCl2

The Attempt at a Solution



show the calculations determining the limiting reactant for each reaction.

Help on how to get the limiting reactant!

And using the limiting reactant as the starting amount, determine the amount of hydrogen gas that was produced per reaction flask:
 
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I think you need to know the molarity of HCl.
 
oh my bad 1.0moles HCL/1L or 0.1 moles HCL/100mL solution
 
Since you know the molarity of HCl to be 1M, you can compute how many moles of HCl you have at 25mL. The formula Molarity = mole/volume in Liter, or mole = Molarity x Volume in liter.
Further, based on your equation, you know you need a 2 HCL to 1 Mg ratio.
So just figure out how many moles of Mg you have at .15g, .3 and .6
From that you can see your limiting reactant.
 
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