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what volume of .250m hcl is needed to react with excess mg in order to form 1.50l of h₂ at 25 c and .950 atm

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Final answer:

To find the volume of 0.250 M HCl needed to react with Mg and form 1.50 L of H2 gas at 25°C and 0.950 atm, apply the ideal gas law to calculate the moles of hydrogen, use stoichiometry to determine moles of HCl, and then divide by the molarity of HCl to find the volume.

Step-by-step explanation:

To calculate the volume of 0.250 M HCl needed to react with excess magnesium (Mg) to form 1.50 L of hydrogen gas (H₂) at 25°C and 0.950 atm, we can use the ideal gas law and the stoichiometry of the reaction between HCl and Mg.

Step 1: Ideal Gas Law to Find Moles of H₂

PV = nRT, where:

After calculating n (moles of H₂), we can use the reaction stoichiometry:

Mg + 2 HCl → MgCl₂ + H₂

Step 2: Stoichiometry to Find Moles of HCl

From the balanced equation, 2 moles of HCl react with Mg to produce 1 mole of H₂. Therefore, we multiply the moles of H₂ by 2 to find the moles of HCl needed.

Step 3: Molarity to Find Volume of HCl

Using the molarity of the HCl solution (0.250 M), we divide the moles of HCl by the molarity to find the necessary volume.

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